Q.Represent the cell and calculate the standard emf of the cell having the following cell reaction: 2Cr(s) + 3Cd^2+(aq) -> 2Cr^3+(aq) + 3Cd(s). Given that, E°(Cr3+/Cr) = -0.73V and E°(Cd2+/Cd) = -0.40V.

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:

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

   …

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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 …

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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). …

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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 …

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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 …

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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). …

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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 …

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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: …

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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.

   …

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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. …

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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): $Cu^{2+} + 2e^- \rightarrow Cu$ …

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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). …

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