Q.The correct cell to represent the following reaction is : $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$ (A) $2Ag | Ag^+ || Zn | Zn^{2+}$ (B) $Ag^+ | Ag || Zn^{2+} | Zn$ (C) $Ag | Ag^+ || Zn | Zn^{2+}$ (D) $Zn | Zn^{2+} || Ag^+ | Ag$

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set ANNUAL1 mark
   Q.What is the potential difference between the two electrodes of the galvanic cell called?

   ›Reveal solutionSolution

   The potential difference between the two electrodes of a galvanic cell (measured when no current is drawn) is called the electromotive force (EMF) or cell potential, $E_{\text{cell}}$.

   Concept. In a galvanic (voltaic) cell, the two half-cells are at different electrode potentials. The difference between the cathode and anode potentials is what pushes electrons through the external circuit:

   $E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}$

   When this potential difference is measured under zero-current conditions (using a potentiometer, so the cell reaction is effectively at equilibrium and no IR drop occurs), it is the maximum potential difference the cell can deliver and is termed the electromotive force (EMF) of the cell.

   ✓Final answer

   It is called the electromotive force (EMF) of the cell — equivalently, the cell potential $E_{\text{cell}}$.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.Consider the following statements about a reaction at equilibrium: A(g) + B(g) ↔ C(g). Statement I: Adding an inert gas at constant volume will shift the equilibrium to the right. Statement II: A catalyst changes the position of equilibrium.

   (a) i) Both statement I and II are correct

   (b) ii) Both statement I and II are incorrect

   (c) iii) Statement I is correct and statement II is incorrect

   (d) iv) Statement I is incorrect and statement II is correct

   ›Reveal solutionSolution

   [!TLDR]

   ii) Both statement I and II are incorrect

   Why

   Adding an inert gas at constant volume does not change partial pressures/concentrations of reacting species, so it does not shift equilibrium (Statement I false). A catalyst speeds up attainment of equilibrium equally in both directions and never shifts its position (Statement II false).

   [!ANSWER]

   ii) Both statement I and II are incorrect

3. CBSE 2025Set ANNUAL1 mark
   Q.For the electrochemical cell $Zn(s)+Cu^{2+}(aq) \rightarrow Zn^{2+}(aq)+Cu(s)$ the cell produces an electrical potential of 1.1 volt, when $[Zn^{2+}]$ and $[Cu^{2+}]$ are unity. State the direction of flow of current on applying external potential of 1.1 volt.

   ›Reveal solutionSolution

   An external potential exactly equal and opposite to the cell's own EMF brings the system to balance, so no net current flows in either direction — this is the basis of potentiometric EMF measurement.

   The Daniell-type cell $Zn(s)|Zn^{2+}(aq)\|Cu^{2+}(aq)|Cu(s)$ spontaneously drives current in the galvanic direction (electrons flow from Zn anode to Cu cathode through the external circuit) with an EMF of 1.1 V under standard conditions.

   If an external opposing potential is applied, it works against this spontaneous cell reaction:

   • If the external potential is less than 1.1 V, the cell's own EMF still dominates, and current continues to flow in the original (galvanic) direction, though at a reduced magnitude.
   • If the external potential is greater than 1.1 V, it overpowers the cell's own EMF, and current is forced to flow in the reverse direction (the cell now behaves as an electrolytic cell, being charged/driven backward).
   • If the external potential exactly equals 1.1 V (as given here), the two opposing potentials exactly cancel, and no net current flows — the system is in electrochemical balance (equilibrium).

   (This exact-balance condition is the working principle behind the potentiometric method of accurately measuring a cell's true EMF.)

   ✓Final answer

   Since the applied external potential (1.1 V) exactly equals and opposes the cell's own EMF, the two cancel and no net current flows in either direction.

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The correct statement in a cell of zinc and copper is

   (a) zinc acts as cathode and copper as anode

   (b) zinc acts as anode and copper as cathode

   (c) the standard reduction potential of zinc is more than that of copper

   (d) the flow of electrons is from copper to zinc

   ›Reveal solutionSolution

   Zinc has a lower (more negative) standard reduction potential than copper, so it is oxidized (anode) while copper is reduced (cathode).

   In a Daniell-type zinc–copper cell, E°(Zn²⁺/Zn) = −0.76 V is lower than E°(Cu²⁺/Cu) = +0.34 V. The electrode with the lower (more negative) reduction potential is oxidized — zinc loses electrons and acts as the anode (Zn → Zn²⁺ + 2e⁻) — while the electrode with the higher reduction potential is reduced — copper gains electrons and acts as the cathode (Cu²⁺ + 2e⁻ → Cu). Electrons flow through the external circuit from zinc to copper (not the reverse), and since Zn has the more negative (smaller) standard reduction potential, statement (c) is false too.

   ✓Final answer

   (b) zinc acts as anode and copper as cathode.

5. CBSE 2024Set D1 markMCQ
   Q.The electromotive force of the cell Zn | ZnSO4 || CuSO4 | Cu is 1.1 volt. Its cathode is

   (a) Zn

   (b) Cu

   (c) ZnSO4

   (d) CuSO4

   ›Reveal solutionSolution

   Reduction happens at the cathode; Cu2+ is reduced to Cu, so Cu is the cathode.

   In the Daniell cell Zn | ZnSO4 || CuSO4 | Cu:

   • Anode (oxidation, left): Zn -> Zn2+ + 2e-
   • Cathode (reduction, right): Cu2+ + 2e- -> Cu

   By convention the electrode written on the right of a cell notation is the cathode where reduction occurs. The standard EMF = E(cathode) - E(anode) = 0.34 - (-0.76) = +1.10 V, matching the given 1.1 V. Because copper has the higher (more positive) reduction potential, Cu2+ is reduced and copper is the cathode.

   ✓Final answer

   (b) Cu — the copper electrode is the cathode (reduction of Cu2+).

6. CBSE 2024Set ANNUAL1 markMCQ
   Q.An electrochemical cell can behave like an electrolytic cell when _______.

   (a) Ecell = 0

   (b) Ecell > Eext

   (c) Eext > Ecell

   (d) Ecell = Eext

   ›Reveal solutionSolution

   A galvanic (electrochemical) cell starts behaving like an electrolytic cell when an external potential greater than the cell's own emf is applied against it, reversing the direction of current flow.

   Consider a Daniell cell: Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s), which normally works as a galvanic cell producing a cell potential Ecell, with electrons flowing from Zn (anode) to Cu (cathode) through the external circuit.

   If an external opposing emf (Eext) is applied to this cell:

   • When Eext < Ecell, the cell continues to work as a galvanic cell, but the current decreases.

   • When Eext = Ecell, no current flows through the cell (this is used to measure the cell's emf accurately, e.g. using a potentiometer).

   • When Eext > Ecell, the direction of current flow reverses. Electrons are now forced to flow from Cu to Zn, so Cu is oxidised and Zn2+ is reduced — exactly the reverse of the spontaneous cell reaction. The cell now behaves as an electrolytic cell, consuming electrical energy to drive a non-spontaneous reaction (this is the working principle of charging a rechargeable/secondary cell such as a lead storage battery).

   ✓Final answer

   The cell behaves as an electrolytic cell when Eext > Ecell (option c).

7. CBSE 2024Set ANNUAL1 mark
   Q.Write True/False: A hydrogen bridge is used to maintain continuity of ion flow in a Daniell cell.

   ›Reveal solutionSolution

   This statement is FALSE. A Daniell cell uses a salt bridge (e.g. containing KCl or KNO3 in agar-agar gel), not any "hydrogen bridge", to complete the internal circuit.

   A Daniell cell consists of a Zn electrode dipped in ZnSO4 solution (anode) and a Cu electrode dipped in CuSO4 solution (cathode), connected externally by a wire and internally by a salt bridge. The salt bridge allows ions to migrate between the two half-cells, maintaining electrical neutrality in each compartment as the cell reaction proceeds, and completes the internal circuit without letting the two solutions mix directly. There is no such thing as a "hydrogen bridge" performing this role.

   ✓Final answer

   False -- it is a salt bridge, not a hydrogen bridge, that maintains continuity of ion flow in a Daniell cell.

8. CBSE 2023Set 56/1/11 markMCQ
   Q.The correct cell to represent the following reaction is : $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$ (A) $2Ag | Ag^+ || Zn | Zn^{2+}$ (B) $Ag^+ | Ag || Zn^{2+} | Zn$ (C) $Ag | Ag^+ || Zn | Zn^{2+}$ (D) $Zn | Zn^{2+} || Ag^+ | Ag$

   ›Reveal solutionSolution

   By convention the anode (oxidation) is written on the left and the cathode (reduction) on the right. Zinc is oxidised and silver ions are reduced, so the cell is $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — option (D).

   A cell diagram is written anode (left) $\parallel$ cathode (right), with each half-cell running from the electrode metal outward and the double bar $\parallel$ marking the salt bridge.

   For the reaction

   $Zn + 2Ag^+ \rightarrow Zn^{2+} + 2Ag$

   • Zinc loses electrons: $Zn \rightarrow Zn^{2+} + 2e^-$ (oxidation, anode, left).
   • Silver ions gain electrons: $Ag^+ + e^- \rightarrow Ag$ (reduction, cathode, right).

   Writing the anode as metal $\mid$ ion and the cathode as ion $\mid$ metal gives

   $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$

   Checking the options:

   • (A) $2Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — electrodes reversed and coefficients are never used.
   • (B) $Ag^+ \mid Ag \parallel Zn^{2+} \mid Zn$ — anode/cathode reversed.
   • (C) $Ag \mid Ag^+ \parallel Zn \mid Zn^{2+}$ — anode/cathode reversed.
   • (D) $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$ — Zn (anode) on the left, Ag (cathode) on the right. Correct.

   ✓Final answer

   The correct representation is (D) $Zn \mid Zn^{2+} \parallel Ag^+ \mid Ag$.

9. CBSE 2022Set E1 markMCQ
   Q.The standard electrode potentials for the following reactions are given ( At 25°C ): Ag+(aq) + e- -> Ag(s), E° Ag+/Ag = +0.80 V ; Sn2+(aq) + 2e -> Sn(s), E° Sn2+/Sn = -0.14 V. The electromotive force (EMF) of the given cell Sn | Sn2+ (1M) || Ag+ (1M) | Ag is

   (a) 0.66 V

   (b) 0.80 V

   (c) 1.08 V

   (d) 0.94 V

   ›Reveal solutionSolution

   For Sn | Sn2+ || Ag+ | Ag, EMF = E°(Ag+/Ag) - E°(Sn2+/Sn) = 0.80 - (-0.14) = 0.94 V.

   In the cell notation the left electrode is the anode (oxidation) and the right is the cathode (reduction):

   • Cathode (reduction): Ag+ + e- -> Ag, E° = +0.80 V
   • Anode (oxidation): Sn -> Sn2+ + 2e-, E°(Sn2+/Sn) = -0.14 V

   E°cell = E°cathode - E°anode = (+0.80) - (-0.14) = +0.94 V.

   (The positive value confirms the reaction Sn + 2Ag+ -> Sn2+ + 2Ag is spontaneous. Standard potentials are intensive, so they are not multiplied by the number of electrons.)

   ✓Final answer

   (d) 0.94 V.

10. CBSE 2022Set ANNUAL1 markMCQ
    Q.Which one of the following statements is incorrect for a voltaic cell ?

    (a) It converts chemical energy to electrical energy.

    (b) It uses electrical energy to carry out chemical changes.

    (c) It is based on a redox reaction.

    (d) It has $-\Delta G$.

    ›Reveal solutionSolution

    A voltaic cell produces electricity from a spontaneous redox reaction — it does not consume electrical energy, so statement (b) describes an electrolytic cell instead.

    Checking each option against what a voltaic (galvanic) cell actually does:

    • (a) True — a voltaic cell converts chemical energy into electrical energy.

    • (b) False — this describes an electrolytic cell, which uses externally supplied electrical energy to force a non-spontaneous chemical change. A voltaic cell does the opposite.

    • (c) True — it is based on a spontaneous redox (oxidation–reduction) reaction.

    • (d) True — a spontaneous cell reaction has $\Delta G < 0$, i.e. $-\Delta G$.

    ✓Final answer

    (b) — "It uses electrical energy to carry out chemical changes" is the incorrect statement for a voltaic cell.

11. CBSE 2022Set ANNUAL1 markMCQ
    Q.For the given cell reaction $Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$:

    (a) $Mg$ as cathode

    (b) $Cu$ as cathode

    (c) $Cu$ is oxidizing agent

    (d) None of the above

    ›Reveal solutionSolution

    By IUPAC convention the electrode written on the LEFT of a cell is the anode and the one on the RIGHT is the cathode. Here Mg is the anode and Cu is the cathode. Option (B).

    The cell is written as $Mg\,|\,Mg^{2+}\,||\,Cu^{2+}\,|\,Cu$.

    Convention: anode (negative, oxidation) is written on the left; cathode (positive, reduction) is written on the right.

    The electrode reactions are:

    • Anode (Mg, oxidation): $Mg \rightarrow Mg^{2+} + 2e^-$
    • Cathode (Cu, reduction): $Cu^{2+} + 2e^- \rightarrow Cu$

    Since $E^\circ_{Mg^{2+}/Mg} = -2.37\ V$ and $E^\circ_{Cu^{2+}/Cu} = +0.34\ V$, Mg (more negative) is oxidised and acts as the anode; $Cu^{2+}$ is reduced, so Cu is the cathode. ($Cu^{2+}$ is the oxidising agent here, not Cu metal, so option C is wrong.)

    ✓Final answer

    (B) $Cu$ acts as the cathode.

12. CBSE 2021Set A1 markMCQ
    Q.Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is

    (a) Weston cell

    (b) Daniel cell

    (c) Calomel cell

    (d) None of these

    ›Reveal solutionSolution

    A zinc-copper galvanic cell with this notation is the Daniell cell.

    The cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) is the Daniell cell, a galvanic (voltaic) cell.

    • At the anode (LHS): Zn(s) → Zn2+ + 2e- (oxidation).
    • At the cathode (RHS): Cu2+ + 2e- → Cu(s) (reduction).
    • Standard EMF = E°(cathode) - E°(anode) = 0.34 - (-0.76) = +1.10 V.

    The Weston cell (standard cell) uses Cd/Hg, and the calomel electrode uses Hg/Hg2Cl2 — neither matches this notation.

    ✓Final answer

    (B) Daniel cell.