Lesson 14.1.2

14.1.2 Cell emf and cell diagrams Quiz: Pearson Edexcel Chemistry, Unit 14

20 questions

Written by Revision Ninja · Last reviewed 11 October 2026

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Lesson 14.1.2, Cell emf and cell diagrams: 20 multiple choice questions for the Pearson Edexcel Chemistry (9CH0), Unit 14: Redox II, written with Revision Ninja.

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The 20 questions

  1. In a cell diagram, which half-cell is written on the left?

    • The oxidation (anode) half-cell, from which electrons flow out
    • The half-cell that contains the salt bridge
    • The reduction (cathode) half-cell, where electrons are gained
    • Whichever half-cell has the higher E-zero value

  2. What does a single vertical line | represent in a cell diagram?

    • A mixture of two different solutions in the same half-cell, separated by a porous plug
    • A phase boundary, such as between a solid electrode and an aqueous solution
    • A salt bridge between two half-cells, which carries electrons from one half-cell to the other directly
    • A wire carrying electrons between the electrodes, which also separates the two solutions completely

  3. What does a double line || represent in a cell diagram?

    • A salt bridge or porous partition that separates the two half-cells
    • A conductor that carries electrons from the anode to the cathode through the external circuit
    • A phase boundary between a solid and a solution, where the electrode reaction takes place
    • The site of the reaction at the electrode surface, where all the ions are formed and mixed

  4. Which equation gives the standard emf of a cell?

    • E-zero cell = E-zero cathode - E-zero anode
    • E-zero cell = E-zero anode + E-zero cathode
    • E-zero cell = E-zero anode - E-zero cathode
    • E-zero cell = E-zero cathode divided by E-zero anode

  5. Which electrode is the cathode in a standard cell?

    • The electrode that is always made of metal
    • The electrode with the more negative E-zero, where reduction occurs
    • The electrode where oxidation occurs
    • The electrode with the more positive E-zero, where reduction occurs

  6. What must be positive for a cell reaction to be feasible under standard conditions?

    • E-zero of the anode only
    • E-zero cell
    • The sum of the two E-zero values
    • E-zero of the cathode only

  7. What is the function of the salt bridge in an electrochemical cell?

    • It allows electrons to flow directly between the metal electrodes without passing through the external circuit
    • It removes the solid products from the solution, which stops the cell from ever reaching equilibrium
    • It supplies hydrogen gas to the electrode so that the reaction can continue at a fixed rate
    • It completes the circuit by allowing ions to flow, keeping the solutions electrically neutral

  8. Using Ni2+/Ni (E-zero -0.25 V) and Ag+/Ag (E-zero +0.80 V), what is the standard emf of the cell Ni | Ni2+ || Ag+ | Ag?

    • +1.05 V
    • -1.05 V
    • +0.25 V
    • +0.55 V

  9. Which cell diagram corresponds to the cell in which Ni is oxidised and Ag+ is reduced?

    • Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)
    • Ni2+(aq) | Ni(s) || Ag(s) | Ag+(aq)
    • Ni(s) | Ag+(aq) || Ni2+(aq) | Ag(s)
    • Ag(s) | Ag+(aq) || Ni2+(aq) | Ni(s)

  10. Using E-zero Fe2+/Fe = -0.44 V and E-zero Cu2+/Cu = +0.34 V, what is the emf of the cell Fe | Fe2+ || Cu2+ | Cu?

    • +0.10 V
    • +0.78 V
    • +0.34 V
    • -0.78 V

  11. For a two-electron cell with E-zero cell = +0.78 V, what is dG-zero for the cell reaction?

    • About -151 kJ mol-1, from dG = -nFE
    • About +151 kJ mol-1, because a positive emf makes dG positive
    • About -0.78 kJ mol-1, because the emf in volts is simply multiplied by the Faraday constant
    • About -75 kJ mol-1, because only one electron is transferred per mole of reaction

  12. A cell has E-zero cell = -0.40 V as written. What does this indicate?

    • The reaction as written is not feasible, so the reverse reaction is spontaneous
    • The reaction is feasible but very slow
    • The reaction is feasible at high temperature only
    • The reaction is feasible because the absolute value of E is always used

  13. In the cell Zn | Zn2+ || Cu2+ | Cu, which electrode is the negative terminal?

    • Zinc, the anode, from which electrons flow out
    • Copper, the cathode, where reduction occurs
    • Both electrodes are positive
    • The salt bridge is the negative terminal

  14. Why does a cell built from two identical half-cells have zero emf?

    • Concentration differences always cancel the emf exactly
    • Both electrodes produce hydrogen gas, which cancels out
    • Both electrodes have the same E-zero
    • The salt bridge blocks all current under standard conditions

  15. In the cell Zn | Zn2+ || Cu2+ | Cu, what is the reaction at the cathode?

    • Zn2+(aq) + 2e- gives Zn(s), a reduction at the cathode
    • 2H+(aq) + 2e- gives H2(g), a reduction at the cathode
    • Cu2+(aq) + 2e- gives Cu(s), a reduction at the positive electrode
    • Zn(s) gives Zn2+(aq) + 2e-, an oxidation at the cathode

  16. Using the half-cells MnO4-/Mn2+ (E-zero +1.51 V) as cathode and Fe3+/Fe2+ (E-zero +0.77 V) as anode, what is the standard emf?

    • +0.23 V
    • +2.28 V
    • +0.74 V
    • -0.74 V

  17. Which expression relates E-zero cell to the equilibrium constant K for a cell with n electrons?

    • ln K = -nF E-zero cell / RT
    • ln K = RT / (nF E-zero cell)
    • K = nF E-zero cell / RT
    • ln K = nF E-zero cell / RT

  18. For a two-electron cell with E-zero cell = +0.50 V at 298 K, what is K approximately?

    • About 1 x 10^2
    • About 0.50
    • About 39
    • About 8 x 10^16

  19. A student calculates E-zero cell = +0.20 V but measures +0.05 V with non-standard concentrations. What is the best explanation?

    • The salt bridge always adds 0.15 V to any measured emf
    • Standard potentials are unaffected by concentration, so the difference is a voltmeter error
    • Electrode potentials depend on concentrations, so non-standard conditions change E from the standard value
    • E-zero values are only valid at 0 K, so the difference is due to temperature alone

  20. What happens to the emf of a cell as its reaction reaches equilibrium?

    • It remains unchanged, because E-zero is a constant
    • It doubles, because electrons accumulate in the external circuit
    • It becomes negative, because products are formed in excess
    • It falls to zero, because dG = 0 and no further net driving force exists

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