Question

If $E^{\circ} _{(cell)} = 3.17 \,V$ for the cell in which the following reaction takes place: $Mg_{(s)} + 2Ag^+(0.0001M) \to Mg^{2+}(0.130M) + 2Ag_{(s)}$
then its $E_{(cell)}$ will be

• A. 3.56 V
• B. 2.96 V
• C. 5.35 V
• D. 7.25 V

Answer 1

Answer: (B)

The cell can be written as
$Mg |Mg^+ (0.130\,M)||Ag^+(0.0001\,M) |Ag$
$E_{(cell)} = E^{\circ}_{(cell)} - \frac{RT}{2F} ln\frac{[Mg^{2+}]}{[Ag^+]^2}$
$ = 3.17\,V - \frac{0.059\,V}{2} log \frac{0.130}{(0.0001)^2}$
$ = 3.17 \,V - 0.21 \,V = 2.96\,V$