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ChemMedAdv 2011 • Paper 2

Cell Potential Calculation Using Nernst Equation Under Given Conditions

Consider the following cell reaction:
2Fe(s)+O2(g)+4H(aq)+→2Fe(aq)2++2H2O(l)E∘=1.67 V2\mathrm{Fe}_{(\mathrm{s})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}^+_{(\mathrm{aq})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})} + 2\mathrm{H}_2\mathrm{O}_{(\mathrm{l})} \quad \mathrm{E}^\circ = 1.67\text{ V}
At [Fe2+]=10−3 M[\mathrm{Fe}^{2+}] = 10^{-3}\text{ M}, P(O2)=0.1 atm\mathrm{P}(\mathrm{O}_2) = 0.1\text{ atm} and pH=3\mathrm{pH} = 3, the cell potential at 25 ∘C25\ ^\circ\mathrm{C} is

Options

A

1.47 V1.47\text{ V}

B

1.77 V1.77\text{ V}

C

1.87 V1.87\text{ V}

D

1.57 V1.57\text{ V}

Correct

Topics & Concepts

ElectrochemistryElectrochemical CellNernst Equation

Step-by-Step Solution

To determine the cell potential at 25 ∘C25\ ^\circ\mathrm{C}, we use the Nernst equation for the given electrochemical cell reaction:

2Fe(s)+O2(g)+4H(aq)+→2Fe(aq)2++2H2O(l)E∘=1.67 V2\mathrm{Fe}_{(\mathrm{s})} + \mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}^+_{(\mathrm{aq})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})} + 2\mathrm{H}_2\mathrm{O}_{(\mathrm{l})} \quad E^\circ = 1.67\text{ V}

1. Determination of the Number of Electrons Transferred (nn)

The half-cell reactions are:

  • Anode (Oxidation):
    2Fe(s)→2Fe(aq)2++4e−2\mathrm{Fe}_{(\mathrm{s})} \rightarrow 2\mathrm{Fe}^{2+}_{(\mathrm{aq})} + 4e^-
  • Cathode (Reduction):
    O2(g)+4H(aq)++4e−→2H2O(l)\mathrm{O}_{2(\mathrm{g})} + 4\mathrm{H}^+_{(\mathrm{aq})} + 4e^- \rightarrow 2\mathrm{H}_2\mathrm{O}_{(\mathrm{l})}

Thus, the total number of electrons transferred in the balanced cell reaction is n=4n = 4.

2. Expression for the Reaction Quotient (QQ)

The reaction quotient for the process is given by:
Q=[Fe2+]2P(O2)⋅[H+]4Q = \frac{[\mathrm{Fe}^{2+}]^2}{P(\mathrm{O}_2) \cdot [\mathrm{H}^+]^4}

Given data:

  • [Fe2+]=10−3 M[\mathrm{Fe}^{2+}] = 10^{-3}\text{ M}
  • P(O2)=0.1 atm=10−1 atmP(\mathrm{O}_2) = 0.1\text{ atm} = 10^{-1}\text{ atm}
  • pH=3  ⟹  [H+]=10−pH=10−3 M\mathrm{pH} = 3 \implies [\mathrm{H}^+] = 10^{-\mathrm{pH}} = 10^{-3}\text{ M}

Substituting these values into the reaction quotient:
Q=(10−3)2(10−1)⋅(10−3)4=10−610−1⋅10−12=10−610−13=107Q = \frac{(10^{-3})^2}{(10^{-1}) \cdot (10^{-3})^4} = \frac{10^{-6}}{10^{-1} \cdot 10^{-12}} = \frac{10^{-6}}{10^{-13}} = 10^7

3. Application of the Nernst Equation

At T=25 ∘CT = 25\ ^\circ\mathrm{C} (298 K298\text{ K}), the Nernst equation is:
E=E∘−0.0591nlog⁡10QE = E^\circ - \frac{0.0591}{n} \log_{10} Q

Substituting the values:
E=1.67−0.05914log⁡10(107)E = 1.67 - \frac{0.0591}{4} \log_{10}(10^7)
E=1.67−0.0591×74E = 1.67 - \frac{0.0591 \times 7}{4}
E=1.67−0.41374=1.67−0.1034≈1.57 VE = 1.67 - \frac{0.4137}{4} = 1.67 - 0.1034 \approx 1.57\text{ V}

Therefore, the cell potential is 1.57 V1.57\text{ V}, which corresponds to option D.

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