Question

The standard Gibbs energy for the given cell reaction in kJ mol^-1 at 298 K is :
\(\mathrm{Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)}\),
\(E^\circ = 2\ \mathrm{V}\) at 298 K
(Faraday's constant, \(F = 96000\ \mathrm{C\ mol^{-1}}\))

• A. 384
• B. \(-192\)
• C. \(-384\)
• D. 192

Hint:

Use \(\Delta G^\circ=-nFE^\circ\) with n = 2 electrons for Zn to Zn²⁺.

Answer 1

The Correct Option is C

The reaction transfers n = 2 electrons (\(\mathrm{Zn \rightarrow Zn^{2+}+2e^-}\)).
\(\Delta G^\circ = -nFE^\circ = -2 \times 96000 \times 2\ \mathrm{J\ mol^{-1}} = -384000\ \mathrm{J\ mol^{-1}} = -384\ \mathrm{kJ\ mol^{-1}}\).
Answer: (3) \(-384\)