Q.Consider the following reaction : $Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$ Given : $E^o_{Ag^+/Ag} = 0.80$ V, $E^o_{Zn^{2+}/Zn} = -0.76$ V, $1\,F = 96500$ C mol$^{-1}$ $\Delta_r G^o$ for the above reaction is : (A) $-301.080$ kJ mol$^{-1}$ (B) $+310.080$ kJ mol$^{-1}$ (C) $-326.070$ kJ mol$^{-1}$ (D) $-375.060$ kJ mol$^{-1}$

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set 56/3/11 markMCQ
   Q.On electrolysis of very dilute aqueous solution of NaCl using platinum electrodes : (A) $H_2$ gas is evolved at anode. (B) Na is produced at cathode. (C) $O_2$ gas is evolved at anode. (D) $H_2$ gas is evolved at cathode.

   ›Reveal solutionSolution

   In very dilute aqueous NaCl with inert Pt electrodes, water’s reduction to $H_2$ at the cathode and water’s oxidation to $O_2$ at the anode outcompete the NaCl reactions. So $H_2$ is produced at the cathode and $O_2$ at the anode — making option (C) and (D) correct.

   Why standard electrode potentials decide the outcome

   Electrolysis is a battle of competing half-reactions. At each electrode, the species that is easier to oxidise (at the anode) or easier to reduce (at the cathode) will react first. “Easier” means having a more positive reduction potential for reduction, or a more negative reduction potential for oxidation (equivalently, a more positive oxidation potential).

   For a very dilute aqueous solution of NaCl, the possible species are:

   • Cathode (reduction): $Na^+$ ions and $H_2O$ molecules.
   • Anode (oxidation): $Cl^-$ ions and $H_2O$ molecules.

   We compare their standard reduction potentials (at 298 K, 1 M concentration, 1 atm pressure). But remember: concentration matters. In very dilute NaCl, $[Cl^-]$ is tiny, which shifts the actual potential of the chlorine half-reaction significantly.

   ---

   Step-by-step reasoning

   1. What happens at the cathode?

   Two reduction half-reactions compete:

   $\begin{aligned} Na^+ + e^- &\rightarrow Na(s) \quad E^\circ = -2.71\ \text{V} \\[4pt] 2H_2O + 2e^- &\rightarrow H_2(g) + 2OH^- \quad E^\circ = -0.83\ \text{V} \end{aligned}$

   The reduction of water to hydrogen gas has a much less negative (i.e., more positive) standard potential. Even though the actual potential for water reduction depends slightly on pH (here neutral to slightly basic), it remains far above $-2.71\ \text{V}$. So water is reduced preferentially.

   Watch out

   A common mistake is to think that because $Na^+$ is present, sodium metal will plate out. But sodium’s reduction potential is so negative that water (even in neutral solution) is reduced first. Sodium metal would instantly react with water anyway — it’s never produced in aqueous electrolysis.

   Result at cathode: $H_2$ gas is evolved. This matches option (D).

   ---

   2. What happens at the anode?

   Two oxidation half-reactions compete (written as reductions for comparison):

   $\begin{aligned} Cl_2(g) + 2e^- &\rightarrow 2Cl^- \quad E^\circ = +1.36\ \text{V} \\[4pt] O_2(g) + 4H^+ + 4e^- &\rightarrow 2H_2O \quad E^\circ = +1.23\ \text{V} \end{aligned}$

   To decide which is easier to oxidise, we look at the reverse reactions. The more negative the reduction potential, the easier it is to oxidise the reduced form. Here, water (or $OH^-$) has a lower reduction potential ($+1.23\ \text{V}$) than chlorine ($+1.36\ \text{V}$). So water should be oxidised more readily — but only if concentrations are equal.

   Tip

   The Nernst equation is the key: for the chlorine half-reaction, $E = E^\circ - \frac{0.059}{2}\log\frac{[Cl^-]^2}{P_{Cl_2}}$. In very dilute NaCl, $[Cl^-]$ is extremely small, making $\log\frac{[Cl^-]^2}{P_{Cl_2}}$ very negative, so $E$ becomes much larger than $+1.36\ \text{V}$. That means chlorine evolution becomes even harder (requires a more positive potential). Meanwhile, the oxygen evolution potential is nearly independent of $[Cl^-]$ (it depends on pH, which stays near neutral). So water oxidation wins decisively.

   Result at anode: $O_2$ gas is evolved. This matches option (C).

   ---

   3. Eliminating the wrong options

   • (A) $H_2$ at anode: Hydrogen is produced at the cathode, not the anode. Anode is where oxidation occurs — hydrogen would need to be oxidised, but there’s no $H_2$ present initially.
   • (B) Na at cathode: As argued, water reduction outcompetes sodium ion reduction. Sodium metal is never formed in aqueous solution.

   ---

   ✓Final answer

   The correct options are (C) and (D): $O_2$ gas is evolved at the anode and $H_2$ gas is evolved at the cathode.

2. CBSE 2026Set 56/2/11 markMCQ
   Q.Consider the following reaction : $Zn(s) + Ag_2O(s) + H_2O(l) \rightarrow Zn^{2+}(aq) + 2Ag(s) + 2OH^-(aq)$ Given : $E^o_{Ag^+/Ag} = 0.80$ V, $E^o_{Zn^{2+}/Zn} = -0.76$ V, $1\,F = 96500$ C mol$^{-1}$ $\Delta_r G^o$ for the above reaction is : (A) $-301.080$ kJ mol$^{-1}$ (B) $+310.080$ kJ mol$^{-1}$ (C) $-326.070$ kJ mol$^{-1}$ (D) $-375.060$ kJ mol$^{-1}$

   ›Reveal solutionSolution

   Zinc is oxidised and silver is reduced, giving $E^o_{cell} = 0.80 - (-0.76) = 1.56$ V with $n = 2$. Then $\Delta_r G^o = -nFE^o_{cell} = -301.080$ kJ mol$^{-1}$, which is option (A).

   The standard Gibbs energy of a cell reaction is linked to its standard cell potential by

   $\Delta_r G^o = -nFE^o_{cell}$

   so we first find $E^o_{cell}$, then $n$, and finally $\Delta_r G^o$.

   1. Identify the electrodes. Zinc is oxidised (anode) and silver is reduced (cathode):

   $\text{Anode:}\quad Zn \rightarrow Zn^{2+} + 2e^-$

   $\text{Cathode:}\quad Ag_2O + H_2O + 2e^- \rightarrow 2Ag + 2OH^-$

   2. Standard cell potential. Using the given reduction potentials,

   $E^o_{cell} = E^o_{cathode} - E^o_{anode} = 0.80 - (-0.76) = 1.56\ \text{V}$

   3. Electrons transferred. Each half-reaction involves 2 electrons, so $n = 2$.

   4. Gibbs energy.

   $\Delta_r G^o = -nFE^o_{cell} = -(2)(96500\ \text{C mol}^{-1})(1.56\ \text{V})$

   $\Delta_r G^o = -301080\ \text{J mol}^{-1} = -301.080\ \text{kJ mol}^{-1}$

   The negative value confirms the reaction is spontaneous, consistent with the positive $E^o_{cell}$.

   ✓Final answer

   $\Delta_r G^o = \boxed{-301.080\ \text{kJ mol}^{-1}}$ — option (A).

3. CBSE 2025Set D1 markMCQ
   Q.The electromotive force of the following cell is: Zn | Zn2+ (1M) || Fe2+ (1M) | Fe, given E°Zn2+|Zn = -0.76 V, E°Fe2+|Fe = -0.44 V

   (a) 1.2 V

   (b) 0.32 V

   (c) -1.2 V

   (d) -0.32 V

   ›Reveal solutionSolution

   E(cell) = E(cathode) - E(anode) = -0.44 - (-0.76) = +0.32 V.

   In the cell notation Zn | Zn2+ || Fe2+ | Fe, zinc is the anode (oxidation, written left) and iron is the cathode (reduction, written right). The standard cell EMF is:

   E(cell) = E(cathode) - E(anode)

   E(cell) = E(Fe2+/Fe) - E(Zn2+/Zn)

   E(cell) = (-0.44) - (-0.76)

   E(cell) = -0.44 + 0.76 = +0.32 V

   The positive value confirms the reaction (Zn + Fe2+ -> Zn2+ + Fe) is spontaneous.

   ✓Final answer

   (B) 0.32 V.

4. CBSE 2025Set A1 mark
   Q.Write True or False: The cell potential is the addition of the electrode potentials (reduction potentials) of the cathode and anode.

   ›Reveal solutionSolution

   Cell potential is obtained by subtracting the anode's reduction potential from the cathode's, not by adding the two reduction potentials.

   The standard cell potential is defined as:

   $E^{\circ}_{cell} = E^{\circ}_{cathode(reduction)} - E^{\circ}_{anode(reduction)}$

   If both electrode potentials are taken as reduction potentials (as the statement specifies), the correct operation is a subtraction (cathode minus anode), not an addition. The 'addition' phrasing is only valid if the anode's contribution is expressed as an oxidation potential (= −reduction potential): then $E_{cell} = E_{oxidation, anode} + E_{reduction, cathode}$, which is mathematically the same subtraction in disguise. Since the statement explicitly calls both 'reduction potentials' and says they are added, it is not correct as worded.

   ✓Final answer

   False.

5. CBSE 2025Set ANNUAL1 mark
   Q.Answer in one word/sentence: Given the standard electrode potentials, arrange these metals in their increasing order of reducting power: K+/K = -2.93 V, Ag+/Ag = 0.80 V, Hg2+/Hg = 0.79 V, Mg2+/Mg = -2.37 V.

   ›Reveal solutionSolution

   Reducing power increases as the standard electrode (reduction) potential becomes more negative, so we simply rank the four E° values from most positive to most negative.

   Given standard reduction potentials:

   $K^+/K = -2.93\ V,\quad Mg^{2+}/Mg = -2.37\ V,\quad Hg^{2+}/Hg = +0.79\ V,\quad Ag^+/Ag = +0.80\ V$

   A more negative (or less positive) standard reduction potential means the metal has a greater tendency to lose electrons (be oxidised) — i.e. it is a stronger reducing agent. Conversely, a metal with a highly positive reduction potential prefers to stay reduced (gain electrons), making it a poor reducing agent (like Ag, a "noble" metal).

   Ranking E° from most positive (weakest reducing agent) to most negative (strongest reducing agent):

   $Ag\,(+0.80\ V) > Hg\,(+0.79\ V) > Mg\,(-2.37\ V) > K\,(-2.93\ V)$

   So reducing power increases in the reverse order:

   $Ag < Hg < Mg < K$

   ✓Final answer

   Increasing order of reducing power: Ag < Hg < Mg < K (potassium is the strongest reducing agent, silver the weakest).

6. CBSE 2025Set ANNUAL1 mark
   Q.Write two applications of electrochemical series.

   ›Reveal solutionSolution

   Electrochemical series: predicts reaction feasibility and metal-displacement reactivity.

   The electrochemical series arranges elements/ions in order of their standard reduction potentials ($E°$). Two common applications:

   1. Predicting feasibility of a redox reaction: a reaction is spontaneous if the species with the higher (more positive) reduction potential is reduced while the species with the lower (more negative) reduction potential is oxidized, i.e. $E°_{cell} = E°_{cathode} - E°_{anode} > 0$.
   2. Predicting displacement reactions: a metal higher up in the reactivity order (more negative $E°$, i.e. a stronger reducing agent) can displace a metal lower in the series from a solution of its salt — e.g. Zn (more negative $E°$) displaces Cu from $CuSO_4$ solution.

   (Other valid applications include calculating standard cell EMF, and predicting the relative strength of oxidizing/reducing agents.)

   ✓Final answer

   1. To predict the feasibility (spontaneity) of a given redox reaction;
   2. To predict which metal can displace another metal from a solution of its salt (relative reactivity of metals).
7. CBSE 2024Set 56/3/11 markMCQ
   Q.During the electrolysis of aqueous NaCl, the cathodic reaction is : (A) Oxidation of $Cl^-$ ion (B) Reduction of $Na^+$ ion (C) Oxidation of $H_2O$ (D) Reduction of $H_2O$

   ›Reveal solutionSolution

   In aqueous NaCl electrolysis, the cathode is where reduction occurs. The competing reductions are $Na^+$ and $H_2O$; water has a much less negative reduction potential, so it is reduced instead of sodium. The correct answer is (D) Reduction of $H_2O$.

   The key to this question lies in understanding Standard Electrode Potentials — the numerical measure of a species’ tendency to gain electrons (be reduced). In electrolysis, the cathode is the negative electrode where reduction happens. When you have an aqueous solution, you must consider all possible reducible species, not just the obvious cation from the salt.

   For aqueous NaCl, the solution contains:

   • $Na^+$ ions (from the salt)
   • $H_2O$ molecules (the solvent)
   • $Cl^-$ ions (from the salt — but these are oxidised at the anode, not reduced at the cathode)

   At the cathode, two reduction reactions compete:

   1. Reduction of $Na^+$:

      $Na^+ + e^- \rightarrow Na(s)$

      Standard reduction potential: $E^\circ = -2.71\ \text{V}$

   2. Reduction of water:

      $2H_2O + 2e^- \rightarrow H_2(g) + 2OH^-$

      Standard reduction potential: $E^\circ = -0.83\ \text{V}$

   Watch out

   A common mistake is to assume that because $Na^+$ is the cation, it must be reduced at the cathode. But the more positive (or less negative) the reduction potential, the easier the reduction. Here, water’s potential ($-0.83\ \text{V}$) is far less negative than sodium’s ($-2.71\ \text{V}$), meaning water is much more readily reduced.

   Now, let’s work through the reasoning step by step.

   1. Identify the cathode process.

      The cathode is the electrode where reduction occurs — gain of electrons. So we look for which species can accept electrons.

   2. List all reducible species in the solution.

      In aqueous NaCl: $Na^+$ ions and $H_2O$ molecules. (The $Cl^-$ ions are already in their lowest oxidation state for a halide; they cannot be reduced further under these conditions — they are oxidised at the anode.)

   3. Compare their reduction potentials.

      • $Na^+ + e^- \rightarrow Na$: $E^\circ = -2.71\ \text{V}$
      • $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$: $E^\circ = -0.83\ \text{V}$

      The more positive (or less negative) the potential, the stronger the oxidising agent — i.e., the more likely it is to be reduced. Since $-0.83 > -2.71$, water is a much stronger oxidising agent than $Na^+$ in this system.

   4. Apply the principle of preferential discharge.

      During electrolysis, the species with the higher (less negative) reduction potential gets reduced first. So water is reduced at the cathode, producing hydrogen gas and hydroxide ions, not sodium metal.

   5. Check the options.

      • (A) Oxidation of $Cl^-$ — this happens at the anode, not the cathode.
      • (B) Reduction of $Na^+$ — thermodynamically unfavourable compared to water.
      • (C) Oxidation of $H_2O$ — oxidation occurs at the anode.
      • (D) Reduction of $H_2O$ — this is exactly what happens.

   Tip

   A quick way to remember: In aqueous electrolysis, if the cation is from a highly reactive metal (Group 1 or 2, like Na, K, Mg, Ca), water is reduced instead. The metal ion stays in solution. This is why electrolysis of aqueous NaCl gives $H_2$ at the cathode, not Na metal.

   For competing reductions at the cathode:

   $\text{The species with the higher (less negative) } E^\circ \text{ is reduced first.}$

   ✓Final answer

   The correct option is (D) Reduction of $H_2O$.

8. CBSE 2024Set ANNUAL1 mark
   Q.In which electrode of a Galvanic cell, oxidation reaction takes place?

   ›Reveal solutionSolution

   The anode of a galvanic cell is where oxidation (electron loss) occurs.

   A Galvanic (voltaic) cell converts the chemical energy of a spontaneous redox reaction into electrical energy, splitting the reaction into two half-cells. The electrode at which oxidation (loss of electrons) takes place is called the anode; in a galvanic cell this is the negative electrode. For example, in the Daniell cell, $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$ occurs at the zinc anode. The electrode at which reduction (gain of electrons) occurs is the cathode, the positive electrode.

   ✓Final answer

   Oxidation occurs at the anode (negative electrode).

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Emf of a cell with Nickel and Copper electrode will be (Given E0 Ni+2/Ni = -0.25 V, E0 Cu2+/Cu = +0.34 V)

   (a) -0.59 V

   (b) +0.59 V

   (c) +0.09 V

   (d) -0.09 V

   ›Reveal solutionSolution

   The electrode with the higher (more positive) standard reduction potential acts as the cathode; the cell EMF is Ecathode - Eanode.

   Given: E-degree(Ni2+/Ni) = -0.25 V, E-degree(Cu2+/Cu) = +0.34 V.

   Since Cu2+/Cu has the higher reduction potential, copper is reduced (cathode) and nickel is oxidised (anode):

   Anode (oxidation): Ni -> Ni2+ + 2e-

   Cathode (reduction): Cu2+ + 2e- -> Cu

   E-degree(cell) = E-degree(cathode) - E-degree(anode) = 0.34 V - (-0.25 V) = 0.59 V

   Since this is positive, the cell reaction as written is spontaneous.

   ✓Final answer

   (b) +0.59 V.

10. CBSE 2023Set 56/1/11 markMCQ
    Q.$\Delta G$ and $E^\circ_{cell}$ for a spontaneous reaction will be : (A) positive, negative (B) negative, negative (C) negative, positive (D) positive, positive

    ›Reveal solutionSolution

    A spontaneous reaction releases free energy ($\Delta G < 0$) and generates a positive cell potential ($E^\circ_{\text{cell}} > 0$); the answer is (C).

    The connection between thermodynamics and electrochemistry rests on a beautiful relationship: the Gibbs free energy change tells us whether a reaction will proceed on its own, while the standard cell potential measures the driving force behind electron flow. For a reaction to be spontaneous, it must release free energy to do useful work—including pushing electrons through a circuit.

    The fundamental bridge between these quantities is:

    $\Delta G^\circ = -nFE^\circ_{\text{cell}}$

    where $n$ is the number of moles of electrons transferred, $F$ is Faraday's constant ($96{,}485 \, \text{C/mol}$), and $E^\circ_{\text{cell}}$ is the standard cell potential.

    The negative sign in this equation is the key. It tells us that a positive cell potential (electrons flowing spontaneously from anode to cathode, releasing energy) corresponds to a negative Gibbs free energy change (energy released, reaction spontaneous). Think of it this way: when a battery drives current through a device, it's doing work on the surroundings, which means the battery's chemical reaction is losing free energy—hence $\Delta G < 0$.

    Now let's apply this to the question:

    1. What does spontaneity require thermodynamically?

       A spontaneous process proceeds without external intervention and releases free energy. The criterion is $\Delta G < 0$ (negative). This is the defining condition—if $\Delta G$ were positive, we'd need to supply energy to make the reaction go.

    2. What does the equation tell us about $E^\circ_{\text{cell}}$?

       Rearranging: $E^\circ_{\text{cell}} = -\dfrac{\Delta G^\circ}{nF}$. Since $n$ and $F$ are always positive, and we've established that $\Delta G^\circ < 0$ for a spontaneous reaction, the negative sign in front flips the inequality: $E^\circ_{\text{cell}} > 0$ (positive).

    3. Physical interpretation

       A positive standard cell potential means the cathode (reduction site) has a higher reduction potential than the anode (oxidation site). Electrons naturally flow "downhill" in potential, from lower to higher reduction potential, generating voltage. This is exactly what happens in a galvanic (voltaic) cell—the spontaneous reaction produces electrical energy.

    Watch out

    Don't confuse the sign convention. A positive $E^\circ_{\text{cell}}$ corresponds to a negative $\Delta G^\circ$ because of the minus sign in the relationship. Students sometimes flip this, thinking "positive energy" means "positive $\Delta G$," but thermodynamically, releasing energy means losing free energy (negative $\Delta G$).

    Tip

    Remember the mnemonic: Spontaneous = Supplies energy. A spontaneous electrochemical reaction supplies electrical energy (positive voltage) to the external circuit, which means it's releasing Gibbs free energy (negative $\Delta G$).

    Checking the options:

    • (A) positive $\Delta G$, negative $E^\circ_{\text{cell}}$: non-spontaneous reaction
    • (B) negative $\Delta G$, negative $E^\circ_{\text{cell}}$: contradicts $\Delta G^\circ = -nFE^\circ_{\text{cell}}$
    • (C) negative $\Delta G$, positive $E^\circ_{\text{cell}}$: consistent with spontaneity ✓
    • (D) positive $\Delta G$, positive $E^\circ_{\text{cell}}$: contradicts the fundamental equation

    ✓Final answer

    The correct option is (C): $\Delta G$ is negative and $E^\circ_{\text{cell}}$ is positive for a spontaneous reaction.

11. CBSE 2023Set 56/3/11 markMCQ
    Q.Assertion (A): Electrolysis of aqueous solution of NaCl gives chlorine gas at anode instead of oxygen gas. Reason (R): Formation of oxygen gas at anode requires overpotential. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

    ›Reveal solutionSolution

    In the electrolysis of aqueous NaCl, chlorine is produced at the anode instead of oxygen because the overpotential for oxygen evolution makes the actual potential needed for oxygen formation higher than that for chlorine, even though the standard potential for oxygen is lower. Both Assertion and Reason are true, and the Reason correctly explains the Assertion — so the answer is (A).

    Why this question is about real-world electrochemistry

    Standard electrode potentials tell you which reaction is thermodynamically favoured. But electrolysis happens under kinetic conditions. The key twist here: oxygen evolution at an inert anode (like platinum or graphite) has a large overpotential — an extra voltage needed to overcome the activation barrier. Chlorine evolution, on the other hand, has a much smaller overpotential. So the reaction that actually occurs at the anode is not the one with the lower standard potential, but the one that requires the lower actual voltage (standard potential + overpotential).

    Let’s see the numbers.

    ---

    1. What are the possible anode reactions?

    In aqueous NaCl, the solution contains these ions:

    $\text{Na}^+$, $\text{Cl}^-$, $\text{H}^+$ (from water), and $\text{OH}^-$ (from water).

    At the anode, oxidation happens. The two candidates are:

    • Oxidation of chloride ions:

    $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^- \quad E^\circ = +1.36\ \text{V}$

    • Oxidation of water (to oxygen):

    $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^- \quad E^\circ = +1.23\ \text{V}$

    Note

    Standard potentials are given as reduction potentials. For oxidation, we reverse the sign. But when comparing which oxidation is easier, we compare the actual potentials needed — the more negative the oxidation potential (or the lower the reduction potential), the easier it is to oxidise. Here, water oxidation has $E^\circ = +1.23\ \text{V}$ (reduction), so its oxidation potential is $-1.23\ \text{V}$. Chloride oxidation has $E^\circ = +1.36\ \text{V}$ (reduction), so its oxidation potential is $-1.36\ \text{V}$. Since $-1.23 > -1.36$, water oxidation is thermodynamically easier — it should occur first.

    So why doesn’t it?

    ---

    2. The role of overpotential

    Overpotential ($\eta$) is the extra voltage beyond the thermodynamic value required to drive a reaction at a noticeable rate. For oxygen evolution on common anode materials (Pt, graphite), $\eta$ is substantial — typically around $0.4$–$0.6\ \text{V}$. For chlorine evolution on the same materials, $\eta$ is very small (often $<0.1\ \text{V}$).

    So the actual potential needed for each reaction is:

    • For oxygen:

    $E_{\text{actual}}(\text{O}_2) = 1.23\ \text{V} + \eta_{\text{O}_2} \approx 1.23 + 0.5 = 1.73\ \text{V}$

    • For chlorine:

    $E_{\text{actual}}(\text{Cl}_2) = 1.36\ \text{V} + \eta_{\text{Cl}_2} \approx 1.36 + 0.05 = 1.41\ \text{V}$

    Now compare: chlorine requires a lower actual voltage (1.41 V) than oxygen (1.73 V). So chlorine is produced preferentially.

    Watch out

    A common mistake is to compare only the standard potentials and conclude oxygen should form. That would be correct only if overpotentials were zero. In real electrolysis, overpotential flips the order.

    ---

    3. Why the Reason is correct

    The Reason states: “Formation of oxygen gas at anode requires overpotential.” This is true — and it is precisely this overpotential that makes oxygen evolution harder than chlorine evolution under these conditions. Without it, oxygen would be the product. So the Reason is the correct explanation for the Assertion.

    Tip

    Overpotential is why many electrolysis predictions based solely on $E^\circ$ fail. Always ask: “Is there a significant overpotential for one of the products?” For gases on inert electrodes, oxygen almost always has a high overpotential; hydrogen does too, but less so.

    ---

    4. Evaluating the options

    • (A) Both true, Reason explains Assertion — this matches our analysis.
    • (B) Both true, but Reason does not explain — false, because overpotential is the very reason.
    • (C) Assertion true, Reason false — false, Reason is true.
    • (D) Assertion false — false, Assertion is true (chlorine is produced at the anode in practice).

    ✓Final answer

    The correct option is (A) — Both Assertion and Reason are true, and the Reason correctly explains the Assertion.

12. CBSE 2023Set ANNUAL1 markMCQ
    Q.When concentration of Zn2+ and Cu2+ ions is unity (1 mol dm-3), then electrical potential of Daniell cell will be -

    (a) 0.00 V

    (b) 1.10 V

    (c) 1.35 V

    (d) 2.00 V

    ›Reveal solutionSolution

    A Daniell cell is Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s); its EMF is the difference of the two standard reduction potentials, and at unit concentration this IS the standard cell potential.

    The Daniell cell has the cell reaction Zn(s) + Cu2+(aq) -> Zn2+(aq) + Cu(s), with Zn as the anode (oxidation) and Cu as the cathode (reduction).

    Standard reduction potentials: E-standard(Cu2+/Cu) = +0.34 V, E-standard(Zn2+/Zn) = -0.76 V.

    E-cell-standard = E-cathode-standard - E-anode-standard = 0.34 - (-0.76) = 1.10 V.

    Since both Zn2+ and Cu2+ are at unit concentration (1 mol dm^-3), the Nernst equation's log term is zero (log 1 = 0), so the actual cell potential equals the standard potential.

    ✓Final answer

    (b) 1.10 V.