Chemistry
63 Solved Past Papers
2010 – 2024 Archives
Electrochemistry Past Papers
Solved past paper MCQs for Electrochemistry from official UHS, NUMS, SZABMU, DUHS, and KMU examinations. Includes verified distractor autopsies and step-by-step cognitive explanations.
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Oxidation number of 'Mn' in \( KMnO_4 \) is: [UHS 2024]
A
0
B
1
C
-7
D
7
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
For a neutral chemical compound, the algebraic sum of all the individual oxidation numbers must equal zero.
Formula:
$$+1 + x + 4(-2) = 0$$
Solution:
Potassium (K) is a Group 1 metal and always has an oxidation state of \( +1 \).
Oxygen (O) typically has an oxidation state of \( -2 \).
Let \( x \) be the oxidation state of Manganese (Mn).
Set up the equation: \( 1 + x - 8 = 0 \).
Solve for \( x \): \( x - 7 = 0 \implies x = +7 \).
Why other options are incorrect:
Manganese can exist in \( +2, +3, +4, +6 \) states, but in the highly oxidized permanganate ion, it is stripped of 7 valence electrons, resting at \( +7 \).
Which step is irrelevant with respect to balancing of redox equations by oxidation number method? [UHS 2024]
A
Split the reaction into two half reactions
B
Assign oxidation number to all the atoms involved in the equation
C
Identify the element undergoing a change in oxidation number
D
Equalize the number of electrons lost and gained
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
There are two distinct methodologies for balancing redox equations: the Oxidation Number Method and the Ion-Electron (Half-Reaction) Method.
Solution:
In the Oxidation Number Method, the entire skeleton equation is kept intact. You identify the changes in oxidation states and cross-multiply coefficients directly within the single equation to balance the total electron transfer.
Splitting the equation into two physically separate equations (an oxidation half and a reduction half) is the defining, exclusive first step of the Ion-Electron Method.
Therefore, splitting is entirely irrelevant to the oxidation number method.
Why other options are incorrect:
Options B, C, and D are mandatory, foundational steps for executing the oxidation number method.
Which of the following is NOT a correct feature of electrolytic cells? [UHS 2024]
A
Reduction occurs at cathode
B
Oxidation occurs at anode
C
Alternating current source is connected to electrodes
D
Electrochemical reaction takes place
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
An electrolytic cell uses an external power supply to drive a non-spontaneous chemical reaction by forcing electrons in a single, specific direction.
Solution:
Because specific half-reactions must occur at designated electrodes (oxidation at the anode, reduction at the cathode), the polarity of the electrodes must remain strictly constant.
This requires a Direct Current (DC) source, like a standard battery.
If an Alternating Current (AC) source were used, the polarity of the electrodes would rapidly switch back and forth (e.g., 50 times a second), causing the redox reactions to constantly reverse themselves and resulting in zero net electrolysis.
Why other options are incorrect:
Options A, B, and D are fundamental, true characteristics of all electrochemical cells.
During electrolysis of concentrated aqueous solution of NaCl, Which ion is NOT discharged at cathode? [SZABMU-RC 2024]
A
\( Na^+ \)
B
\( Cl^- \)
C
\( H^+ \)
D
\( OH^- \)
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
During the electrolysis of aqueous salts, multiple ions compete. At the cathode, the cation with the higher standard reduction potential wins the competition and is discharged.
Solution:
In an aqueous NaCl solution, two cations migrate to the cathode: \( Na^+ \) (from the salt) and \( H^+ \) (from the water).
\( H^+ \) has a much higher reduction potential than \( Na^+ \).
Consequently, \( H^+ \) is successfully reduced and discharged as \( H_2 \) gas.
\( Na^+ \) fails to reduce and remains dissolved in the solution as a spectator ion.
Why other options are incorrect:
While \( Cl^- \) and \( OH^- \) are technically not discharged at the cathode (they migrate to the anode), the question targets the specific competing cation that fails to discharge. Therefore, \( Na^+ \) is the intended chemical answer.
Which one of the following is a strong electrolyte in solution? [ETEA 2024]
A
Acetic acid
B
Ammonium hydroxide
C
Carbonic acid
D
Potassium iodide
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A strong electrolyte is a substance that dissociates nearly 100% into free, mobile ions when dissolved in water.
Solution:
Potassium Iodide (\( KI \)) is an ionic salt.
Because it is highly soluble, it completely shatters its crystal lattice in water to yield independent \( K^+ \) and \( I^- \) ions.
This massive presence of free ions allows the solution to conduct electricity extremely well, categorizing it as a strong electrolyte.
Why other options are incorrect:
Acetic acid and carbonic acid are weak covalent acids that only partially ionize. Ammonium hydroxide is a weak base that also only partially dissociates.
The electrode potential of the standard hydrogen electrode is chosen as: [ETEA 2024]
A
-1V
B
0V
C
1V
D
2V
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
To measure the absolute potential of a single half-cell, it must be compared against a universal thermodynamic baseline.
Solution:
The global scientific community universally agreed to use the Standard Hydrogen Electrode (SHE) as this baseline.
By IUPAC convention, the Standard Electrode Potential (\( E^\circ \)) of the SHE is arbitrarily assigned an exact value of \( 0.00 \text{ V} \) at all temperatures.
All other standard reduction potentials on the electrochemical series are measured relative to this zero point.
Why other options are incorrect:
Using non-zero integers like \( 1 \text{ V} \) or \( -1 \text{ V} \) would needlessly complicate the mathematics of cell potential calculation.
Fuel cell is a typical galvanic cell which is based on reaction between: [DUHS 2024]
A
Hydrogen and copper
B
Methane and oxygen
C
Hydrogen and oxygen
D
Nitrogen and oxygen
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A fuel cell continuously converts the chemical energy of a fuel and an oxidizing agent directly into electricity, without combustion.
Solution:
The most prevalent, highly efficient, and historically significant type of fuel cell (used extensively in aerospace missions like Apollo) is the Alkaline Fuel Cell (AFC).
This specific cell operates on the continuous redox reaction between Hydrogen gas (the fuel, oxidized at the anode) and Oxygen gas (the oxidant, reduced at the cathode).
The only exhaust byproduct is pure water.
Why other options are incorrect:
While methane can be used in Solid Oxide Fuel Cells, basic textbook fuel cells fundamentally teach the Hydrogen-Oxygen reaction. Copper and Nitrogen are not standard fuel cell reagents.
Zn rod acts as cathode when coupled with magnesium electrode. This is because the reduction potential of: [NUMS 2024]
A
Zn > Mg
B
Fe is precipitated out
C
Cu and Fe both dissolve
D
No reaction taken place
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In a galvanic cell, the electrode composed of the metal with the higher (more positive/less negative) standard reduction potential acts as the cathode.
Solution:
The standard reduction potential of Zinc (\( Zn \)) is \( -0.76 \text{ V} \).
The standard reduction potential of Magnesium (\( Mg \)) is \( -2.37 \text{ V} \).
Because \( -0.76 \text{ V} \) is mathematically greater than \( -2.37 \text{ V} \) (Zn > Mg), Zinc has a stronger thermodynamic pull on electrons.
Zinc forces Magnesium to oxidize (act as the anode), while Zinc undergoes reduction (acts as the cathode).
Why other options are incorrect:
Note: Options B, C, and D are physical misprints in the original source book, accidentally copied from the subsequent question. The only chemically relevant and correct answer is A.
Placing a rod of iron metal in a solution of \( CuSO_4 \): [NUMS 2024]
A
Cu will be deposited
B
Fe is precipitated out
C
Cu and Fe both dissolve
D
No reaction taken place
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A single displacement redox reaction occurs when a more reactive solid metal is placed into a solution containing ions of a less reactive metal.
Solution:
Look at the electrochemical series. Iron (Fe) has a reduction potential of \( -0.44 \text{ V} \), while Copper (Cu) has a reduction potential of \( +0.34 \text{ V} \).
Because Iron is higher in the reactivity series (it is a stronger reducing agent), it wants to lose its electrons to the Copper ions.
The Iron rod will begin to dissolve (oxidize) into \( Fe^{+2} \) ions: \( Fe \rightarrow Fe^{+2} + 2e^- \).
The \( Cu^{+2} \) ions in the blue sulfate solution will accept those electrons and reduce into solid Copper metal: \( Cu^{+2} + 2e^- \rightarrow Cu \).
This solid Copper will be deposited on the iron rod and at the bottom of the beaker.
Why other options are incorrect:
Iron does not precipitate; it actively dissolves. Copper does not dissolve; it precipitates out of solution. The reaction is highly spontaneous, so "no reaction" is false.
In redox reaction of \( SO_2 \) with \( KMnO_4 \) in acidic medium: [ETEA 2023]
A
\( SO_2 \) oxides \( KMnO_4 \)
B
\( SO_2 \) reduces \( KMnO_4 \)
C
\( KMnO_4 \) is inert in acidic medium
D
\( SO_2 \) can't undergo redox reaction
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Potassium Permanganate (\( KMnO_4 \)) is a famous, incredibly strong oxidizing agent, especially in acidic environments. Sulfur dioxide (\( SO_2 \)) often acts as a reducing agent.
In \( KMnO_4 \), Manganese is in a highly unstable \( +7 \) oxidation state and desperately wants to acquire electrons.
In \( SO_2 \), Sulfur is in a \( +4 \) state, but can easily be oxidized to the \( +6 \) state (sulfate).
\( SO_2 \) provides the electrons to \( KMnO_4 \), causing Manganese to be reduced from \( +7 \) to \( +2 \) (evidenced by the deep purple solution turning colorless).
Because \( SO_2 \) forces \( KMnO_4 \) to gain electrons, \( SO_2 \) acts as the reducing agent.
Why other options are incorrect:
\( SO_2 \) does not oxidize \( KMnO_4 \); it does the exact opposite. \( KMnO_4 \) is highly active in acidic media, not inert.
Zinc displaces copper from its solution because: [ETEA 2023]
A
Atomic no. of zinc is higher than that of copper
B
Zinc has higher reduction potential than copper
C
Zinc is more soluble
D
Zinc has smaller reduction potential than copper
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A metal can spontaneously displace another metal from its aqueous salt solution if it acts as a stronger reducing agent.
Solution:
The strength of a reducing agent is inversely proportional to its reduction potential. A smaller (more negative) reduction potential means a higher tendency to lose electrons (oxidize).
Zinc has a standard reduction potential of \( -0.76 \text{ V} \).
Copper has a standard reduction potential of \( +0.34 \text{ V} \).
Because Zinc has a smaller reduction potential than Copper, solid Zinc will naturally oxidize, forcing the Copper ions in solution to reduce and precipitate out.
Why other options are incorrect:
Option B is factually inverted (Zinc's potential is lower, not higher). Atomic number and general solubility are not the primary thermodynamic drivers for redox displacement.
In electrochemical series, what is the electrode potential of all metals above Hydrogen? [DUHS 2023]
A
Zero
B
Negative
C
Positive
D
Greater than zero
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The standard electrochemical series is anchored around the Standard Hydrogen Electrode (SHE).
Solution:
The SHE is assigned a standard reduction potential of exactly \( 0.00 \text{ V} \).
Metals placed "above" hydrogen in the series (like Lithium, Zinc, Iron) are stronger reducing agents than Hydrogen gas.
Because they have a greater tendency to undergo oxidation rather than reduction compared to Hydrogen, their standard reduction potentials are negative (less than zero).
Why other options are incorrect:
Only Hydrogen has a potential of exactly zero. Metals below Hydrogen (like Copper, Silver, Gold) have positive (greater than zero) reduction potentials.
When aqueous solution of NaCl is electrolyzed; [BUMHS 2023]
A
\( Cl_2 \) is evolved at the cathode
B
\( H_2 \) is evolved at cathode
C
Na is deposited at the cathode
D
Na appears at the anode
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
During the electrolysis of an aqueous solution, multiple cations migrate to the cathode. The cation with the higher (less negative) reduction potential is preferentially reduced.
Solution:
An aqueous solution of NaCl contains \( Na^+ \) ions from the salt and \( H^+ \) ions from the autoionization of water.
The standard reduction potential of \( Na^+ \) to Na is \( -2.71 \text{ V} \).
The reduction potential for the formation of \( H_2 \) gas from water is roughly \( -0.83 \text{ V} \).
Because \( -0.83 \text{ V} \) is much higher than \( -2.71 \text{ V} \), the Hydrogen ions are preferentially reduced, and \( H_2 \) gas rapidly evolves at the cathode.
Why other options are incorrect:
Sodium metal is too reactive to exist in water; it would immediately react even if it did form. Chlorine gas (\( Cl_2 \)) evolves at the anode, not the cathode.
During electrolysis, reduction always occurs at the: [NUMS 2023]
A
Anode
B
Cathode
C
SHE
D
Salt bridge
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
The nomenclature for electrochemical electrodes is strictly tied to the type of chemical reaction occurring at their surface.
Solution:
By universal IUPAC definition, the cathode is the electrode where reduction (gain of electrons) takes place.
The anode is the electrode where oxidation (loss of electrons) takes place.
This holds true whether the cell is galvanic (spontaneous) or electrolytic (forced via an external power source). The mnemonic "RED CAT" (Reduction at Cathode) is always reliable.
Why other options are incorrect:
Oxidation occurs at the anode. The salt bridge is a pathway for spectator ions, not a site of electron transfer.
In the reaction \( Cr_2O_7^{-2} + 14H^+ \rightarrow 2Cr^{+3} + 7H_2O \)
How many electrons are gained or lost by chromium atom? [SZABMU 2023, 2024]
A
+12 e^-
B
+3 e^-
C
+6 e^-
D
-6 e^-
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
To determine electron transfer per atom, calculate the change in the oxidation state of that specific element from reactant to product.
Solution:
First, find the oxidation state of Chromium (Cr) in the dichromate ion (\( Cr_2O_7^{-2} \)): \( 2x + 7(-2) = -2 \implies 2x - 14 = -2 \implies 2x = +12 \implies x = +6 \).
On the product side, Chromium is \( Cr^{+3} \), so its oxidation state is \( +3 \).
The oxidation state of a single Chromium atom goes from \( +6 \) down to \( +3 \).
This mathematical reduction of 3 indicates a gain of 3 electrons (\( +3e^- \)) per Chromium atom.
Why other options are incorrect:
The entire dichromate molecule (which contains two Cr atoms) gains a total of 6 electrons. However, the question specifically asks for the transfer "by chromium atom" (singular), which is 3.
An electrochemical cell is based upon which type of reaction? [UHS 2022]
A
Acid-base reaction
B
Nuclear reaction
C
Redox reaction
D
Neutralization reaction
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Electrochemistry is the study of electricity relating to chemical reactions, specifically relying on the physical transfer of electrons between chemical species.
Solution:
The entire foundation of both Galvanic (voltaic) and Electrolytic cells is the spatial separation of electron loss (oxidation) and electron gain (reduction).
Because these cells operate strictly via oxidation and reduction half-reactions, the fundamental underlying mechanism is a Reduction-Oxidation (Redox) reaction.
Why other options are incorrect:
Acid-base (neutralization) reactions involve the transfer of protons (\( H^+ \)), not electrons, and do not inherently generate electric current. Nuclear reactions involve changes to the atomic nucleus, not valence shell electrons.
In which of the following, Oxygen shows fractional oxidation number? [UHS 2022]
A
\( OF_2 \)
B
\( KO_2 \)
C
\( Na_2O_2 \)
D
\( Cl_2O_7 \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
While Oxygen usually has an oxidation state of \( -2 \), its state changes when bonded to more electronegative halogens, or when forming peroxides and superoxides.
Solution:
Let's determine the oxidation state of O in Potassium superoxide, \( KO_2 \).
Potassium (K) is a Group 1 alkali metal and strictly holds a \( +1 \) oxidation state.
Set up the equation: \( +1 + 2x = 0 \implies 2x = -1 \implies x = -1/2 \).
The fractional oxidation state of Oxygen in a superoxide is \( -1/2 \).
Why other options are incorrect:
In \( OF_2 \), Oxygen is \( +2 \) (since Fluorine is always \( -1 \)). In Sodium peroxide (\( Na_2O_2 \)), Oxygen is \( -1 \). In Dichlorine heptoxide (\( Cl_2O_7 \)), Oxygen is its normal \( -2 \).
Which statement correctly describes the term Standard Electrode potential? [SZABMU 2022]
A
It is the electrode potential determined at room temperature and pressure
B
It is the electrode potential determined under standard condition using Hydrogen Electrode as the other electrode
C
It is the electrode potential of an element and its solution compare to zinc electrode
D
It is the potential which is measured when two half cells are connected
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Single electrode potentials cannot be measured in isolation; they must be measured as a potential difference against a universally agreed-upon standard reference.
Solution:
The scientific community defines the Standard Hydrogen Electrode (SHE) as the ultimate reference point (assigned \( 0.00 \text{ V} \)).
The term "Standard Electrode Potential" (\( E^\circ \)) specifically means coupling the target half-cell (at standard conditions: \( 1 \text{ M} \) concentration, \( 1 \text{ atm} \) pressure, \( 298 \text{ K} \)) against the SHE and recording the voltage on a voltmeter.
Why other options are incorrect:
Option A mentions temperature and pressure but neglects the critical requirement of a reference electrode. Option C incorrectly suggests Zinc is the universal reference. Option D is simply the definition of cell potential, not a standard single electrode potential.
Oxidation number of an element in free state is: [SZABMU 2022]
A
Negative
B
Positive
C
Zero
D
±1
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Oxidation numbers keep track of electron distribution in compounds. If an element has not formed bonds with any other unique element, there has been no electron transfer.
Solution:
By fundamental IUPAC definition, any element existing in its uncombined, pure, free state (whether monoatomic like \( He \), diatomic like \( O_2 \), or polyatomic like \( S_8 \)) has not gained or lost any electrons relative to its neutral atomic state.
Therefore, its assigned oxidation number is exactly Zero.
Why other options are incorrect:
Positive and negative oxidation states only arise when an atom forms bonds with an atom of a different element with differing electronegativity.
The branch of science which deals with the conversion of electrical energy into chemical energy and vice versa is called: [SZABMU 2022]
A
Electrochemistry
B
Thermochemistry
C
Stereochemistry
D
Biochemistry
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The interaction between electrical current and chemical reactions is a massive, dedicated subset of physical chemistry.
Solution:
Electrochemistry encompasses two main phenomena:
1. Galvanic/Voltaic processes: Spontaneous chemical reactions generating electrical energy (e.g., batteries).
2. Electrolytic processes: Using external electrical energy to force non-spontaneous chemical reactions to occur (e.g., electroplating).
Why other options are incorrect:
Thermochemistry studies heat/energy changes in reactions. Stereochemistry studies the 3D spatial arrangement of atoms. Biochemistry studies chemical processes within living organisms.
A cathode has the reduction potential: [ETEA 2022]
A
Less than the anode
B
More than the anode
C
Same as that of anode
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In any standard electrochemical cell, the half-cell with the stronger "pull" on electrons will force the other half-cell to give them up.
Solution:
Reduction potential is literally a measure of a species' tendency to acquire electrons and undergo reduction.
The cathode is defined strictly as the electrode where reduction occurs.
Therefore, to spontaneously drive electrons from the anode to the cathode, the cathode must possess a higher (more positive) reduction potential than the anode.
Why other options are incorrect:
If the cathode had a lesser reduction potential, it would act as the anode instead. If they were the same, no potential difference would exist and the cell voltage would be zero.
The stronger the reduction potential, the more difficult it is to: [ETEA 2022]
A
Reduce the compound
B
Oxidize the compound
C
Electrolyze the compound
D
Neither reduce oxides the compound
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Reduction potential and oxidation potential are mathematically equal but opposite in sign. \( E^\circ_{red} = -E^\circ_{ox} \).
Solution:
A high, strong (very positive) reduction potential means a species desperately wants to gain electrons (it wants to be reduced).
Because it holds onto its electrons tightly, it strongly resists giving them away.
Giving away electrons is the definition of oxidation. Therefore, a high reduction potential makes the species highly resistant and difficult to oxidize.
Why other options are incorrect:
A strong reduction potential makes it mathematically very easy to reduce the compound, not difficult.
The electrolysis of dilute acid solution gives: [DUHS 2022]
A
Cl2 at cathode & H2 at anode
B
H2 at cathode & Cl2 at anode
C
H2 at cathode & O2 at anode
D
H2 at anode & Na at cathode
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Electrolysis of a dilute acid (such as dilute \( H_2SO_4 \)) is essentially the electrolysis of water, facilitated by the acid's ions which increase electrical conductivity.
Solution:
At the cathode (reduction site): The abundant \( H^+ \) ions from the acid are easily reduced to form Hydrogen gas. \( 2H^+ + 2e^- \rightarrow H_2 \).
At the anode (oxidation site): Water molecules (or \( OH^- \) ions) are oxidized because the counter anions (like \( SO_4^{-2} \)) are highly stable and very difficult to oxidize compared to water. \( 2H_2O \rightarrow O_2 + 4H^+ + 4e^- \).
Therefore, \( H_2 \) evolves at the cathode, and \( O_2 \) evolves at the anode.
Why other options are incorrect:
Chlorine gas would only evolve if you electrolyzed concentrated hydrochloric acid or brine, not a generic dilute acid. Hydrogen always evolves at the cathode (since it is a positive ion), never the anode.
A strong electrolyte is one that ____ Completely in solution. [DUHS 2022]
A
Reacts
B
Decomposes
C
Disappears
D
Ionizes
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Electrolytes are substances that produce an electrically conducting solution when dissolved in a polar solvent, such as water.
Solution:
The strength of an electrolyte is strictly defined by the extent to which it dissociates into independent, mobile ions.
A strong electrolyte (such as \( NaCl \) or \( HCl \)) breaks apart practically 100% into its constituent cations and anions when placed in water.
This process of forming ions is called ionization or dissociation.
Why other options are incorrect:
Decomposition implies the permanent breakdown of chemical bonds into entirely different elemental substances. "Disappears" is non-scientific. Ionization accurately describes the physical separation into charged particles.
Highly reactive alkali metals cannot be extracted from aqueous solutions via electrolysis because water would be reduced instead of the metal ions.
Solution:
To extract pure metallic Sodium, industrial chemists use a specialized electrolytic apparatus known as the Downs cell.
In the Downs cell, molten Sodium Chloride (\( NaCl \), mixed with \( CaCl_2 \) to lower the melting point) is electrolyzed.
Sodium ions (\( Na^+ \)) are reduced at the iron cathode to form liquid metallic Sodium, while Chloride ions (\( Cl^- \)) are oxidized at the carbon anode to form Chlorine gas.
Why other options are incorrect:
Magnesium is typically extracted using a Dow cell (similar, but specifically for \( MgCl_2 \)). Aluminum uses the Hall-Héroult process. Iron is extracted via pyrometallurgy in a blast furnace.
Potential of standard hydrogen electrode (SHE) is arbitrarily taken as: [NUMS 2022]
A
10
B
-1
C
+1
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
Just as sea level is assigned an arbitrary altitude of 0 meters to measure the height of mountains, electrochemistry requires a "sea level" to measure standard voltages.
Solution:
The Standard Hydrogen Electrode (SHE) consists of platinum submerged in \( 1 \text{ M} \) \( H^+ \) solution, with \( H_2 \) gas bubbled at \( 1 \text{ atm} \).
By universal international convention, the thermodynamic value for the reduction potential (and oxidation potential) of this specific half-cell is set exactly to Zero Volts (\( 0.00 \text{ V} \)) at all temperatures.
Why other options are incorrect:
Values like \( +1 \) or \( -1 \) are not used. A baseline of exactly zero allows for a mathematically simple reference scale featuring positive and negative values.
During electrolysis, reduction always occurs at: [NUMS 2022]
A
Anode
B
SHE
C
Cathode
D
Salt bridge
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The definitions of the electrodes are tied strictly to the chemical processes that occur at their surfaces, regardless of whether the cell is galvanic or electrolytic.
Solution:
A universal mnemonic in electrochemistry is RED CAT (Reduction occurs at the Cathode) and AN OX (Oxidation occurs at the Anode).
In an electrolytic cell, the battery pumps electrons into the cathode. Positive ions (cations) migrate to the cathode to gain these electrons, completing the process of reduction.
Why other options are incorrect:
Oxidation occurs at the anode. A salt bridge is an internal pathway for ion exchange, not a site for redox reactions.
The chemical reactivity of a metal is tied to its willingness to lose electrons (oxidation potential). Metals with very high positive reduction potentials are extremely resistant to oxidation.
Solution:
Gold (Au) sits at the very bottom of the standard electrochemical series.
It has an exceptionally high standard reduction potential (\( Au^{+3} / Au \approx +1.50 \text{ V} \)).
This implies that Gold has virtually zero tendency to spontaneously lose electrons to the environment, making it a highly inert, noble metal.
Why other options are incorrect:
Scandium and Zinc have highly negative reduction potentials, making them very reactive. Copper is less reactive than Zn but far more reactive than Gold.
Which of the following is strong reducing agent? [PMC 2021]
A
Li
B
F
C
Na
D
Mg
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
A reducing agent donates electrons. The strongest reducing agents are elements that most desperately want to lose an electron, which translates to having the most negative standard reduction potential.
Solution:
Lithium (Li) sits at the absolute top of the standard electrochemical series.
It has the lowest (most negative) standard reduction potential of all elements (\( -3.05 \text{ V} \)).
Because its oxidation potential is incredibly high, solid Lithium is the strongest reducing agent in aqueous conditions.
Why other options are incorrect:
Fluorine (F) sits at the absolute bottom of the series and is the strongest oxidizing agent. While Sodium (Na) and Magnesium (Mg) are strong reducing agents, they are thermodynamically weaker than Lithium.
Electrolysis of a dilute solution of NaCl results at the anode: [PMC 2021]
A
Sodium
B
Chlorine
C
Hydrogen
D
Oxygen
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
In the electrolysis of aqueous solutions, competition occurs at the electrodes between the ions from the salt and the water molecules themselves.
Solution:
In a dilute aqueous solution of NaCl, the concentration of \( Cl^- \) ions is very low compared to water molecules.
At the anode (where oxidation occurs), two reactions compete: the oxidation of \( Cl^- \) to \( Cl_2 \) gas, and the oxidation of \( H_2O \) (or \( OH^- \)) to \( O_2 \) gas.
Thermodynamically, the oxidation of water (requiring \( +1.23 \text{ V} \)) is easier than the oxidation of chloride ions (requiring \( +1.36 \text{ V} \)).
Because the solution is dilute, the overpotential effect that usually favors Chlorine gas in concentrated brine is negligible. Thus, Oxygen gas is evolved at the anode.
Why other options are incorrect:
Chlorine gas is only evolved if the NaCl solution is highly concentrated (brine). Hydrogen is evolved at the cathode, not the anode.
Which of the following is feasible reaction? [PMC 2021]
A
\( Pb + SnO \rightarrow PbO + Sn \)
B
\( Pb + Zn \rightarrow PbO + Zn \)
C
\( Zn + HCl \rightarrow ZnCl_2 + H_2 \)
D
\( 2Ag + 2HCl \rightarrow 2AgCl + H_2 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A single displacement reaction between a metal and an acid is feasible if the metal has a negative standard reduction potential (placing it above Hydrogen in the electrochemical series).
Solution:
Zinc (Zn) has a standard reduction potential of \( -0.76 \text{ V} \).
Because it is "above" Hydrogen (\( 0.00 \text{ V} \)), it acts as a stronger reducing agent.
Zinc will spontaneously donate electrons to the \( H^+ \) ions in Hydrochloric Acid, dissolving into \( ZnCl_2 \) and liberating \( H_2 \) gas. This makes option C highly feasible.
Why other options are incorrect:
Silver (Ag) in option D is a coinage metal situated below Hydrogen (positive reduction potential); it cannot displace Hydrogen from acids. Options A and B are thermodynamically non-spontaneous based on their relative reduction potentials.
The oxidation state of "S" in the \( S_2O_3^{-2} \) is: [NMDCAT 2020]
A
+4
B
+6
C
-2
D
+2
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The average oxidation state of an element in a polyatomic ion is calculated by ensuring the sum of all oxidation states equals the net ionic charge.
Formula:
$$2x + 3(-2) = -2$$
Solution:
Let \( x \) be the average oxidation state of Sulfur (S) in the thiosulfate ion (\( S_2O_3^{-2} \)).
The oxidation state of Oxygen (O) is \( -2 \).
Set up the algebraic equation: \( 2x + 3(-2) = -2 \).
Solve for \( x \): \( 2x - 6 = -2 \implies 2x = +4 \implies x = +2 \).
Why other options are incorrect:
Sulfur has an oxidation state of \( +6 \) in sulfates (\( SO_4^{-2} \)) and \( +4 \) in sulfites (\( SO_3^{-2} \)). Thiosulfate specifically yields an average of \( +2 \).
The common oxidation number of halogens is: [NMDCAT 2020]
A
-1
B
+1
C
-2
D
0
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Halogens are Group 17 elements (Fluorine, Chlorine, Bromine, Iodine) possessing seven valence electrons.
Solution:
To achieve a stable noble gas electron configuration (an octet), a halogen atom requires only one additional electron.
Due to their high electronegativity, they readily pull one electron from metals or less electronegative non-metals.
By gaining one electron, they predominantly form halide ions (e.g., \( F^-, Cl^-, Br^- \)) with an oxidation state of \( -1 \).
Why other options are incorrect:
While halogens (except Fluorine) can exhibit positive oxidation states (like \( +1, +3, +5, +7 \)) when bonded to more electronegative oxygen, the most common and dominant state across all halogens is \( -1 \).
Stronger is the oxidizing agent, stronger is the: [NUMS 2019]
A
emf of cell
B
Oxidation potential
C
Reduction potential
D
Redox potential
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
An oxidizing agent causes another substance to be oxidized while it itself is reduced.
Solution:
The strength of an oxidizing agent is directly proportional to its tendency to gain electrons.
The thermodynamic measure of the tendency to gain electrons is the Standard Reduction Potential (\( E^\circ_{red} \)).
Therefore, the stronger the oxidizing power of a chemical species, the more positive and higher its standard reduction potential.
Why other options are incorrect:
A strong oxidation potential indicates a strong reducing agent, not an oxidizing agent. EMF refers to the total voltage of a complete cell, not a single species. Redox potential is a generic term.
Which of the following metal does not liberate hydrogen on reaction with acid? [NUMS 2019]
A
Mg
B
Pt
C
Zn
D
Ca
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
Metals located above Hydrogen in the electrochemical series (negative reduction potentials) are reactive enough to displace \( H^+ \) ions from dilute acids, generating \( H_2 \) gas.
Solution:
Platinum (Pt) is a highly unreactive, noble metal situated far below Hydrogen in the electrochemical series.
It has a high positive reduction potential (\( +1.20 \text{ V} \)).
Because its oxidation potential is negative relative to Hydrogen, it cannot spontaneously reduce \( H^+ \) ions into \( H_2 \) gas.
Why other options are incorrect:
Magnesium, Zinc, and Calcium all have negative reduction potentials (they lie above Hydrogen) and act as strong reducing agents, eagerly dissolving in acids to liberate hydrogen gas.
Fuel cells require an electrolyte to facilitate the internal transport of ions between the anode and the cathode while acting as an electrical insulator to force electrons through the external circuit.
Solution:
The traditional alkaline fuel cell (AFC), famous for its use in the Apollo space missions, utilizes an aqueous solution of Potassium Hydroxide (\( KOH \)).
The \( KOH \) solution allows for the rapid mobility of hydroxide (\( OH^- \)) ions from the cathode to the anode, which is necessary to sustain the redox reactions.
Why other options are incorrect:
Sodium chloride and sodium nitrate are not used because they do not provide the necessary \( OH^- \) ions or \( H^+ \) ions required for the standard hydrogen-oxygen half-reactions.
The oxidation state of nitrogen in \( NH_4NO_3 \) are: [ETEA 2019]
A
-3 and 5
B
+5 and 3
C
-3 and -3
D
Zero
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
Ammonium nitrate is an ionic compound composed of two distinct polyatomic ions: the ammonium cation (\( NH_4^+ \)) and the nitrate anion (\( NO_3^- \)). The nitrogen atoms in these two ions exist in completely different oxidation states.
Solution:
For the Ammonium ion (\( NH_4^+ \)): Let \( x \) be the oxidation state of N. \( x + 4(+1) = +1 \implies x = -3 \).
For the Nitrate ion (\( NO_3^- \)): Let \( y \) be the oxidation state of N. \( y + 3(-2) = -1 \implies y - 6 = -1 \implies y = +5 \).
Therefore, the oxidation states of the two nitrogen atoms are \( -3 \) and \( +5 \), respectively.
Why other options are incorrect:
If one erroneously tried to average the nitrogens as \( N_2H_4O_3 \), they might guess something else, but chemically, the ions are distinct and must be calculated separately.
A disproportionation reaction is a specific type of redox reaction where the exact same element is simultaneously oxidized and reduced.
Solution:
Look at the oxidation state of Iodine (I) on the reactant side: \( I_2 \) is in its elemental state, so its oxidation number is \( 0 \).
On the product side, Iodine exists in two distinct compounds.
In Sodium Iodate (\( NaIO_3 \)): \( +1 + x + 3(-2) = 0 \implies x = +5 \) (Oxidation).
In Sodium Iodide (\( NaI \)): \( +1 + x = 0 \implies x = -1 \) (Reduction).
Because Iodine is both oxidized and reduced simultaneously, this is definitively a redox reaction.
Why other options are incorrect:
There is no insoluble solid formed (so not precipitation). There are no free radicals involved. It is an internal electron transfer, not a simple physical substitution.
Keeping in view the values of standard reduction potential given above, which one of the following would you select as a feasible redox chemical reaction? [MDCAT 2019]
A
\( 2Au + 6H^+ \rightarrow 2Au^{3+} + 3H_2 \)
B
\( 2Cl^- + I_2 \rightarrow Cl_2 + 2I^- \)
C
\( Mg + 2H^+ \rightarrow Mg^{2+} + H_2 \)
D
\( Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A spontaneous (feasible) redox reaction occurs when a metal with a highly negative reduction potential is oxidized while transferring its electrons to a species with a less negative (or positive) reduction potential.
Solution:
Let's evaluate Option C: Magnesium solid reacting with Hydrogen ions.
Magnesium has a very negative reduction potential (\( -2.37 \text{ V} \)), making it an extremely strong reducing agent (it wants to oxidize).
Hydrogen has a higher reduction potential (\( 0.00 \text{ V} \)).
Therefore, Mg will easily displace Hydrogen from acidic solutions, undergoing oxidation while \( H^+ \) is reduced to \( H_2 \) gas. The cell potential is vastly positive (\( +2.37 \text{ V} \)), making it highly feasible.
Why other options are incorrect:
In option D, Cu (\( +0.34 \text{ V} \)) cannot reduce \( Zn^{2+} \) (\( -0.76 \text{ V} \)). Options A and B include elements (Au, Cl, I) whose potentials are not even listed, though scientifically Au cannot displace H+, and I2 cannot oxidize Cl-.
The potential difference of an electrochemical cell is measured by: [MDCAT 2018]
A
Galvanometer
B
Calorimeter
C
Voltmeter
D
Ammeter
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Electrical potential difference, also known as voltage or electromotive force (EMF), is a measure of the driving force pushing electrons through a circuit.
Solution:
A Voltmeter is specifically designed with high internal resistance to measure the potential difference between two points in a circuit without drawing significant current itself.
In electrochemistry, it connects the anode and cathode to measure the cell's voltage.
Why other options are incorrect:
A galvanometer detects the presence and direction of minute currents. An ammeter measures the magnitude of electrical current (flow of charge), not potential. A calorimeter measures heat changes in thermodynamic reactions.
The standard electrode potential of hydrogen is arbitrarily taken at 298k is ____. [MDCAT 2018]
A
1.00 volt
B
0.10 volt
C
0.00 volt
D
10.0 volt
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
Standard reduction potentials cannot be measured in absolute terms; they require a baseline reference electrode to calculate relative potential differences.
Solution:
The scientific community universally agreed to use the Standard Hydrogen Electrode (SHE) as this baseline.
By IUPAC convention, the standard electrode potential of SHE is assigned an arbitrary, exact value of \( 0.00 \text{ V} \) at all temperatures (including standard \( 298 \text{ K} \)).
Why other options are incorrect:
Values like \( 1.00 \text{ V} \) or \( 0.10 \text{ V} \) are simply random numbers. The entire thermodynamic table of standard potentials revolves around Hydrogen being exactly zero.
In \( NO_3^- \), the oxidation number of "N" is: [MDCAT 2017]
A
+5
B
+3
C
+2
D
-3
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The sum of the oxidation states of all atoms in a polyatomic ion must equal the net charge of the ion.
Formula:
$$x + 3(-2) = -1$$
Solution:
Let \( x \) be the oxidation state of Nitrogen (N).
The oxidation state of Oxygen (O) is generally \( -2 \).
Set up the algebraic equation: \( x + 3(-2) = -1 \).
Solve for \( x \): \( x - 6 = -1 \implies x = +5 \).
Why other options are incorrect:
A value of \( -3 \) would apply to Ammonia (\( NH_3 \)), not Nitrate. Values of \( +3 \) or \( +2 \) correspond to nitrites or nitrogen monoxide, respectively.
The E° value of standard copper half-cell is +0.34V, measured when it is connected with SHE i.e. Standard hydrogen electrode. In this case the half reaction taking place at SHE is: [MDCAT 2017]
A
\( 2H^+_{(aq)} + 2e^- \rightarrow H_{2(g)} \)
B
\( H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^- \)
C
\( 2H^+_{(aq)} + 2e^- \rightarrow 2H_{2(g)} \)
D
\( H_{2(g)} \rightarrow 2H_{(g)} + 2e^- \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
When two half-cells are connected, the one with the higher (more positive) standard reduction potential acts as the cathode (undergoes reduction), and the other acts as the anode (undergoes oxidation).
Solution:
Copper has a standard reduction potential of \( +0.34 \text{ V} \), which is greater than that of SHE (\( 0.00 \text{ V} \)).
Because Copper has a higher reduction potential, it forces the SHE to act as the anode.
At the anode, oxidation (loss of electrons) occurs.
The oxidation half-reaction for hydrogen gas is: \( H_{2(g)} \rightarrow 2H^+_{(aq)} + 2e^- \).
Why other options are incorrect:
Option A represents reduction, which would only happen if SHE were connected to a metal with a negative reduction potential (like Zn). Option C is stoichiometrically incorrect. Option D shows the formation of atomic hydrogen gas instead of aqueous \( H^+ \) ions.
During space flights, astronauts obtained water from: [MDCAT 2017]
A
Nickel cadmium cells
B
Lead accumulator
C
Fuel Cell
D
Alkaline battery
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
A hydrogen-oxygen fuel cell continuously converts the chemical energy of a fuel (\( H_2 \)) and an oxidant (\( O_2 \)) directly into electrical energy.
Solution:
In a standard Apollo-era fuel cell, Hydrogen is oxidized at the anode and Oxygen is reduced at the cathode.
The overall highly exothermic cell reaction is: \( 2H_2 + O_2 \rightarrow 2H_2O \).
One of the most significant advantages of this specific power source in aerospace engineering is that its only chemical byproduct is pure, perfectly potable liquid water, which astronauts utilize for drinking.
Why other options are incorrect:
Nickel-cadmium, lead-acid accumulators, and alkaline batteries are closed-system storage cells that do not produce harvestable drinking water as a byproduct.
$$10\text{Cl}^- + 16\text{H}^+ + 2\text{MnO}_4^- \longrightarrow 5\text{Cl}_2 + 2\text{Mn}^{2+} + 8\text{H}_2\text{O}$$ Which statement is true about this reaction?
A
Manganese is oxidized from +7 to +2.
B
Chlorine is reduced from zero to -1.
C
Chloride ions are reduced from -1 to zero.
D
Manganese is reduced from +7 to +2.
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
In a redox reaction, oxidation corresponds to an increase in oxidation number (loss of electrons), whereas reduction corresponds to a decrease in oxidation number (gain of electrons).
Manganese: In the permanganate ion (\(\text{MnO}_4^-\)), let the oxidation number of \(\text{Mn}\) be \(x\): $$x + 4(-2) = -1 \implies x = +7$$ In the product \(\text{Mn}^{2+}\), the oxidation number is \(+2\). Since the oxidation state decreases from \(+7 \to +2\), Manganese is reduced (it gains 5 electrons).
Chlorine: Reactant is \(\text{Cl}^-\) (oxidation state \(-1\)) and product is \(\text{Cl}_2\) (oxidation state \(0\)). Since the oxidation state increases from \(-1 \to 0\), Chloride ions are oxidized (loss of electrons).
Why other options are incorrect:
Opt_A: Manganese undergoes a decrease in oxidation number (\(+7 \to +2\)), which is reduction, not oxidation.
Opt_B: Chlorine in the reactants is present as \(\text{Cl}^-\) (state \(-1\)), not zero. The forward reaction oxidizes \(-1 \to 0\).
Opt_C: Going from \(-1 \to 0\) is an algebraic increase in oxidation number (loss of \(e^-\)), which defines oxidation, not reduction.
Coinage metals Cu, Ag and Au are the least reactive because they have: [MDCAT 2016]
A
Negative reduction potential
B
Negative oxidation potential
C
Positive reduction potential
D
Positive oxidation potential
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
The reactivity of a metal is determined by its tendency to lose electrons (oxidize). Metals with high positive reduction potentials prefer to gain electrons rather than lose them, making them chemically stable or "noble."
Solution:
In the electrochemical series, elements placed below Hydrogen have positive standard reduction potentials (\( E^\circ_{red} > 0 \)).
Coinage metals (Copper, Silver, Gold) all lie below Hydrogen in the series.
A positive reduction potential indicates that they are poor reducing agents and very resistant to oxidation (corrosion), which is why they are ideal for making coins and jewelry.
Why other options are incorrect:
Alkali and alkaline earth metals (like Na and Mg) have negative reduction potentials (and thus highly positive oxidation potentials), making them extremely reactive. Coinage metals are the opposite.
The diagram shows a galvanic cell. The current will flow from:
Galvanic Cell (Electron Flow from Standard Hydrogen Anode to Cu Cathode)
[MDCAT 2016]
A
Hydrogen electrode to copper electrode
B
Copper electrode to hydrogen electrode
C
Hydrogen electrode to HCl solution
D
CuSO4 solution to hydrogen electrode
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In a standard galvanic cell, electrons flow spontaneously from the anode (lower reduction potential) to the cathode (higher reduction potential).
Solution:
The cell pairs a Standard Hydrogen Electrode (\( 0.00 \text{ V} \)) with a Copper half-cell (\( +0.34 \text{ V} \)).
Because Hydrogen has the lower reduction potential, it undergoes oxidation and acts as the anode.
Copper acts as the cathode.
The electron flow (frequently referred to colloquially as "current" in MDCAT nomenclature) originates at the anode and travels through the external wire to the cathode: from the Hydrogen electrode to the Copper electrode.
Why other options are incorrect:
Flowing from Copper to Hydrogen would require a non-spontaneous input of energy (electrolytic cell). Current does not flow from the electrode directly into its own bulk solution in an external circuit context.
In which of the following reaction hydrogen acts as oxidizing agent. [ETEA 2016]
A
\( H_2 + Cl_2 \rightarrow 2HCl \)
B
\( C_2H_4 + H_2 \rightarrow C_2H_6 \)
C
\( 2Na + H_2 \rightarrow 2NaH \)
D
\( N_2 + 3H_2 \rightarrow 2NH_3 \)
View Answer & Propolis Autopsy
Correct Key: Option CDiagnostic Explanation
Concept:
An oxidizing agent removes electrons from another substance, meaning the oxidizing agent itself gets reduced (its oxidation state decreases).
Solution:
We must find the reaction where Hydrogen's oxidation state decreases from \( 0 \) to a negative value.
In option C: \( 2Na + H_2 \rightarrow 2NaH \).
Sodium is a highly electropositive Group 1 metal (it desperately wants to lose an electron).
Elemental Hydrogen (\( 0 \)) is forced to accept an electron from Sodium, becoming the hydride ion (\( H^- \)) with an oxidation state of \( -1 \).
Because Hydrogen gains electrons and is reduced, it acts as the oxidizing agent.
Why other options are incorrect:
In reactions with more electronegative non-metals like Chlorine or Nitrogen, Hydrogen acts as the reducing agent, losing electron density and taking on a \( +1 \) oxidation state.
Chlorine acts as an oxidizing agent and sodium as reducing agent
C
Chloride acts as a reducing agent and sodium as reducing agent
D
None of the above
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
In ionic bond formation between a metal and a non-metal, the metal acts as a reducing agent (loses electrons) and the non-metal acts as an oxidizing agent (gains electrons).
Solution:
In the reactants, both Sodium (\( Na \)) and Chlorine (\( Cl_2 \)) are in their free elemental states with oxidation numbers of \( 0 \).
In the product (\( NaCl \)), Na is \( +1 \) and Cl is \( -1 \).
Sodium's oxidation state increases (\( 0 \rightarrow +1 \)), so it undergoes oxidation. It is the reducing agent.
Chlorine's oxidation state decreases (\( 0 \rightarrow -1 \)), so it undergoes reduction. It is the oxidizing agent.
Why other options are incorrect:
Option A falsely claims Sodium is reduced. Option C is grammatically and scientifically nonsensical. Option B is factually perfect.
A cell is constructed of the following two half cells. What is Emf of the cell?
\( Ag^+ + e^- \rightleftharpoons Ag + 0.80V \) \( Al^{+3} + 3e^- \rightleftharpoons Al - 1.67V \) [ETEA 2016]
A
2.47V
B
0.087
C
-0.87 V
D
5.81 V
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
The standard Electromotive Force (EMF) of a cell is calculated by subtracting the reduction potential of the anode from the reduction potential of the cathode.
A redox reaction is spontaneous if the metal being oxidized sits "higher" (has a more negative reduction potential) in the electrochemical series than the metal ion being reduced.
Solution:
Zinc (\( Zn \)) has a standard reduction potential of \( -0.76 \text{ V} \).
Copper (\( Cu \)) has a standard reduction potential of \( +0.34 \text{ V} \).
Because Zn is a stronger reducing agent than Cu, solid Zinc will naturally oxidize to \( Zn^{2+} \) while forcing \( Cu^{2+} \) to reduce to solid Cu.
This is the classic, highly spontaneous reaction that powers the Daniell cell (yielding \( +1.10 \text{ V} \)).
Why other options are incorrect:
Option B is the electrolysis of molten NaCl, requiring massive external energy. Option C is the reverse of the Daniell cell and is non-spontaneous. Option D is an extremely non-spontaneous decomposition.
A spontaneous redox reaction occurs when a metal with a higher oxidation potential displaces a metal with a lower oxidation potential from its solution.
Solution:
The given values are standard oxidation potentials (notice the reactions show the loss of electrons).
Zinc has a much higher oxidation potential (\( +0.76 \text{ V} \)) than Copper (\( -0.34 \text{ V} \)).
Therefore, solid Zinc will spontaneously oxidize into \( Zn^{+2} \), and aqueous \( Cu^{+2} \) will be forced to undergo reduction into solid Copper.
The spontaneous overall reaction is: \( Zn_{(s)} + Cu^{+2}_{(aq)} \rightarrow Zn^{+2}_{(aq)} + Cu_{(s)} \).
Why other options are incorrect:
Option A represents the reverse (non-spontaneous) reaction. Options C and D show fundamentally impossible redox stoichiometry where both species undergo reduction.
Note on Signs: Standard reduction potentials are \( E^\circ_{\text{red}}(\text{Zn}) = -0.76\text{ V} \) and \( E^\circ_{\text{red}}(\text{Cu}) = +0.34\text{ V} \). Because the half-reactions in the question prompt depict electron loss (oxidation), the values given are standard oxidation potentials (\( +0.76\text{ V} \) and \( -0.34\text{ V} \)).
Keeping in mind the electrode potential, which one of the following reactions is feasible? [MDCAT 2015]
A
\( Zn^{+2} + Cu \rightarrow Cu^{+2} + Zn \)
B
\( Fe + CuSO_4 \rightarrow FeSO_4 + Cu \)
C
\( Zn + MgSO_4 \rightarrow ZnSO_4 + Mg \)
D
\( Cd + MgSO_4 \rightarrow CdSO_4 + Mg \)
View Answer & Propolis Autopsy
Correct Key: Option BDiagnostic Explanation
Concept:
A metal can displace another metal from its aqueous salt solution only if it sits "higher" (has a more negative reduction potential) in the electrochemical series.
Solution:
Iron (Fe) has a standard reduction potential of \( -0.44 \text{ V} \), whereas Copper (Cu) has a potential of \( +0.34 \text{ V} \).
Because Fe is a stronger reducing agent than Cu, solid Iron will easily oxidize and displace Copper ions from the sulfate solution.
The reaction \( Fe + CuSO_4 \rightarrow FeSO_4 + Cu \) is highly spontaneous and feasible.
Why other options are incorrect:
In option A, Cu cannot displace Zn. In option C, Zn cannot displace Mg (Mg is highly reactive). In option D, Cd cannot displace Mg.
In the figure given below, the electron flow in external circuit is from:
Galvanic Cell (Electron Flow from Anode Zn to Cathode Cu)
[MDCAT 2013]
A
Zinc to copper electrode
B
Right to left
C
Copper to zinc electrode
D
porous partition to zinc electrode
View Answer & Propolis Autopsy
Correct Key: Option ADiagnostic Explanation
Concept:
In any standard galvanic (voltaic) cell, electrons always flow through the external wire from the anode (site of oxidation) to the cathode (site of reduction).
Solution:
Zinc (Zn) has a lower standard reduction potential (\( -0.76 \text{ V} \)) compared to Copper (\( +0.34 \text{ V} \)).
Therefore, Zinc acts as the anode and gets oxidized (loses electrons: \( Zn \rightarrow Zn^{+2} + 2e^- \)).
Copper acts as the cathode and gets reduced (gains those electrons).
The electrons travel from the Zinc electrode, through the external circuit, to the Copper electrode.
Why other options are incorrect:
Electrons never flow from a higher reduction potential (Cu) to a lower one (Zn) spontaneously. The porous partition regulates internal ion flow, not external electron flow.
In voltaic cell a salt bridge is used in order to: [MDCAT 2011]
A
Pass the electric current
B
Prevent the flow of ions
C
Mix solutions of two half cells
D
Allow movement of ions between two cells
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
A salt bridge acts as an electrical connection between two half-cells in a galvanic (voltaic) cell, maintaining electrical neutrality.
Solution:
During the operation of a voltaic cell, oxidation occurs at the anode (generating positive ions) and reduction at the cathode (depleting positive ions).
If this charge imbalance is not neutralized, the cell potential quickly drops to zero.
The salt bridge allows the flow of spectator ions between the two half-cells to balance this charge without allowing the bulk solutions to mix.
Why other options are incorrect:
The salt bridge does not pass electric current (electrons travel through the external wire). It does not prevent ion flow, but rather facilitates it. It specifically exists to prevent the physical mixing of the two reactive solutions.
In an electrochemical series, elements are arranged on the basis of: [MDCAT 2010]
A
pH scale
B
pKa scale
C
pOH scale
D
Hydrogen scale
View Answer & Propolis Autopsy
Correct Key: Option DDiagnostic Explanation
Concept:
The electrochemical series is a systematic arrangement of elements based on their standard electrode potentials.
Solution:
All standard electrode potentials are measured relative to a universal reference point.
This reference is the Standard Hydrogen Electrode (SHE), which is arbitrarily assigned a potential of exactly \( 0.00 \text{ V} \).
Therefore, the series is fundamentally arranged on the basis of the "Hydrogen scale" (increasing or decreasing standard reduction potential relative to Hydrogen).
Why other options are incorrect:
pH, pOH, and pKa scales are used to measure the acidity, basicity, or dissociation constants of chemical species in solution, not their standard redox potentials.
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