Question

The standard cell potential of the following cell Zn|Zn2+ (aq)|Fe2+(aq)|Fe is 0.32 V. Calculate the standard Gibbs energy change for the reaction:Zn(s) + Fe2+(aq) → Zn2+(aq) + Fe(s)(Given: 1 F = 96487 C)

• A. −61.75 kJ mol−1
• B. +5.006 kJ mol−1
• C. −5.006 kJ mol−1
• D. +61.75 kJ mol−1

Answer 1

Answer: (A)

−61.75 kJ mol−1

Answer 2

The standard Gibbs energy change ($\Delta G^\circ$) is related to the standard cell potential (E$^\circ$) by:
$\Delta G^\circ = -nFE^\circ$
where:
n is the number of moles of electrons transferred in the balanced redox reaction.
F is Faraday's constant (96487 C mol$^{-1}$).
E$^\circ$ is the standard cell potential.
In the given reaction, Zn(s) is oxidized to Zn$^{2+}$(aq) and Fe$^{2+}$(aq) is reduced to Fe(s).
Thus, n=2. E$^\circ$ = 0.32 V
$\Delta G^\circ = -(2 mol)(96487 C mol^{-1})(0.32 V)$
$\Delta G^\circ = -61751.04 J mol^{-1} \approx -61.75 kJ mol^{-1}$