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

The Correct Option is A

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
\(\Delta G^\circ = -(2 mol)(96487 C mol^{-1})(0.32 V) \)
\(\Delta G^\circ = -61751.04 J mol^{-1} \approx -61.75  kJ mol^{-1}\)