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^: The standard cell potential (or standard electrode potential), measured in volts (V). This is the potential when all reactants and products are in their standard states (1 M concentratio

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0.487 V\text{0.487 V}

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∘^{\circ}: The standard cell potential (or standard electrode potential), measured in volts (V). This is the potential when all reactants and products are in their standard states (1 M concentration for solutions, 1 atm pressure for gases, 298 K temperature).

  • RR: The ideal gas constant, 8.314 J mol−1 K−18.314 \text{ J mol}^{-1} \text{ K}^{-1}.
  • TT: The absolute temperature in Kelvin (K).
  • nn: The number of moles of electrons transferred in the balanced redox reaction.
  • FF: Faraday's constant, 96500 C mol−196500 \text{ C mol}^{-1}.
  • QQ: The reaction quotient. For a generic reaction aA+bB⇌cC+dDaA + bB \rightleftharpoons cC + dD, Q=[C]c[D]d[A]a[B]bQ = \frac{[C]^c [D]^d}{[A]^a [B]^b}. It represents the ratio of product concentrations to reactant concentrations at any given time, raised to the power of their stoichiometric coefficients. Pure solids and liquids are not included in QQ.

1. (c) A zinc rod is placed in a 0.1 M solution of zinc sulphate at 27 °C. Calculate the potential of the electrode at this temperature, assuming 76 % dissociation of zinc sulphate and E∘(Zn2+/Zn)=0.52 VE^{\circ}(\text{Zn}^{2+}/\text{Zn}) = 0.52 \text{ V}.

  1. Write the half-reaction and determine the number of electrons transferred.

    The reduction half-reaction for the zinc electrode is:

    Zn2+(aq)+2e−⇌Zn(s)\text{Zn}^{2+}(aq) + 2e^- \rightleftharpoons \text{Zn}(s)

    From this, the number of moles of electrons transferred is n=2n = 2.

  2. Calculate the actual concentration of Zn2+\text{Zn}^{2+} ions.

    Given that the initial concentration of ZnSO4\text{ZnSO}_4 is 0.1 M0.1 \text{ M} and its dissociation is 76%76\%:

    [Zn2+]=0.1 M×0.76=0.076 M[\text{Zn}^{2+}] = 0.1 \text{ M} \times 0.76 = 0.076 \text{ M}

  3. Convert the temperature to Kelvin.

    Given temperature T=27 °CT = 27 \text{ °C}.

    T=27+273=300 KT = 27 + 273 = 300 \text{ K}

  4. Apply the Nernst equation for the half-cell.

    The Nernst equation is E=E∘−RTnFln⁡QE = E^{\circ} - \frac{RT}{nF} \ln Q.
    For the reduction of Zn2+\text{Zn}^{2+} to Zn(s)\text{Zn}(s), the reaction quotient Q=1[Zn2+]Q = \frac{1}{[\text{Zn}^{2+}]} (since Zn(s)\text{Zn}(s) is a pure solid).
    So, the Nernst equation becomes:

    E=E∘−RTnFln⁡(1[Zn2+])E = E^{\circ} - \frac{RT}{nF} \ln \left(\frac{1}{[\text{Zn}^{2+}]}\right) E=E∘+RTnFln⁡[Zn2+]E = E^{\circ} + \frac{RT}{nF} \ln [\text{Zn}^{2+}]

  5. Substitute the given values into the Nernst equation.

    Given:
    E∘(Zn2+/Zn)=0.52 VE^{\circ}(\text{Zn}^{2+}/\text{Zn}) = 0.52 \text{ V}
    R=8.314 J mol−1 K−1R = 8.314 \text{ J mol}^{-1} \text{ K}^{-1}
    T=300 KT = 300 \text{ K}
    n=2n = 2
    F=96500 C mol−1F = 96500 \text{ C mol}^{-1}
    [Zn2+]=0.076 M[\text{Zn}^{2+}] = 0.076 \text{ M}

    E=0.52 V+(8.314 J mol−1 K−1)(300 K)(2)(96500 C mol−1)ln⁡(0.076)E = 0.52 \text{ V} + \frac{(8.314 \text{ J mol}^{-1} \text{ K}^{-1})(300 \text{ K})}{(2)(96500 \text{ C mol}^{-1})} \ln (0.076)

  6. Calculate the value.

    First, calculate the term RTnF\frac{RT}{nF}:

    RTnF=(8.314)(300)(2)(96500)=2494.2193000≈0.012923 V\frac{RT}{nF} = \frac{(8.314)(300)}{(2)(96500)} = \frac{2494.2}{193000} \approx 0.012923 \text{ V}

    Next, calculate ln⁡(0.076)\ln(0.076):

    ln⁡(0.076)≈−2.5770\ln(0.076) \approx -2.5770

    Now, substitute these values back into the equation for EE:

    E=0.52 V+(0.012923 V)(−2.5770)E = 0.52 \text{ V} + (0.012923 \text{ V})(-2.5770) E=0.52 V−0.03331 VE = 0.52 \text{ V} - 0.03331 \text{ V} E=0.48669 VE = 0.48669 \text{ V}

    Rounding to three significant figures:

    E≈0.487 VE \approx 0.487 \text{ V}

    The potential of the electrode is 0.487 V\text{0.487 V}.

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