Question

The Gibbs energy change (in J) for the given reaction at [Cu²⁺] = [Sn²⁺] = 1 M and 298 K is :

Cu(s) + Sn²⁺(aq.) $\longrightarrow$ Cu²⁺ (aq.) + Sn(s);

($E_{Sn^{2+}|Sn}^0$ = –0.16 V, $E_{Cu^{2+}|Cu}^0$ = 0.34 V, Take F = 96500 C mol^–1)

Numerical Value Type

Comments

Electrochemistry, Gibbs free energy

Answer 1

Ans. (96500)

Sol. $E_{cell}^0=E_{Sn^{2+}/Sn}^0-E_{Cu^{2+}/Cu}^0$

= – 0.16 –0.34

= – 0.50V

$\Delta G^0 =-nF\ E_{cell}^0$

= – 2 × 96500 × (–0.5)

= 96500J

= 96.5 KJ = 96500 J