Question

The standard Gibbs energy for the given cell reaction in kJ mol^-1 at 2'98 K is:

\[\ce{Zn(s) + Cu^{2+} (aq) -> Zn^{2+} (aq) + Cu(s)}\]

E° = 2 V at 298 K

(Faraday's constant, F = 96000 C mol^-1)

पर्याय

• - 384

• 384

• 192

• - 192

Answer 1

- 384

Explanation:

`Delta"G"^circ = - "nFE"_"cell"^circ`

= - 2 × (96000) × 2 V = - 384000 J/mol = - 384 kJ/mol