Q.(a)

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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):

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

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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 $n = 2$. …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.In the following reaction 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3, which statement is correct?

   (a) 'P' is oxidised only

   (b) 'P' is reduced only

   (c) 'P' is oxidised as well as reduced

   (d) 'P' is neither oxidised nor reduced

   ›Reveal solutionSolution

   When the same element, starting from a single oxidation state, ends up in both a higher and a lower oxidation state among the products, that is a disproportionation reaction.

   Reaction: 4P + 3KOH + 3H2O -> 3KH2PO2 + PH3

   Oxidation state of P in elemental phosphorus (P4, written here as P): 0 (element in its standard state).

   Oxidation state of P in KH2PO2 (potassium hypophosphite): Using K = +1, H = +1 (bonded to O, standard H), O = -2:

   (+1) + 2(+1 for the two H bonded to O) ... more directly: for the hypophosphite ion H2PO2-, charge = -1. With 2 H at +1 and 2 O at -2: 2(+1) + x + 2(-2) = -1 => 2 + x - 4 = -1 => x = +1.

   So P is +1 in KH2PO2 -- this is an INCREASE from 0, i.e. P is OXIDISED here.

   Oxidation state of P in PH3: H bonded to P (a less electronegative element than H here, since P and H have similar/P slightly higher electronegativity by the modified scale used for this compound) is taken as -1 in this hydride convention: x + 3(-1) = 0 => x = +3?

   …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.In the disproportionation reaction 2 Cu⁺ (aq) ⇌ Cu (s) + Cu²⁺ (aq), the cuprous ion, Cu⁺

   (a) undergoes reduction only

   (b) undergoes both reduction and oxidation

   (c) undergoes oxidation only

   (d) does not undergo redox

   ›Reveal solutionSolution

   Disproportionation means the same species is simultaneously oxidised and reduced; here half the Cu⁺ goes to Cu (reduction) and half to Cu²⁺ (oxidation) — option (B).

   Disproportionation is a redox reaction in which an element in one intermediate oxidation state is simultaneously oxidised and reduced.

   Track the oxidation number of copper in $2\,Cu^{+} \rightarrow Cu + Cu^{2+}$:

   • One $Cu^{+}$ (oxidation state $+1$) → $Cu$ (oxidation state $0$): gain of electron = reduction. …

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5. 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) …

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6. 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 …

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7. 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).

   …

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8. 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. $E°_{cell} = E°_{cathode} - E°_{anode} > 0$. …

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9. 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. …

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10. 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, $Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-$ o …

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11. 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

    …

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12. 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. …

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