Question

For the cell $Mg|M{g}^{2+}||A{g}^{2+}|Ag$. Calculate the equilibrium constant at ${25}^{\circ }C$ and the maximum work that can be obtained during the operation of the cell.
(Given: $E_{Mg/M{g}^{2+}}^{\circ }=2.37volt$ and $E_{A{g}^{2+}/Ag}^{\circ }=0.80volt$).

• A. $K=4.86\times {10}^{107};W_{max}=5.118\times {10}^{5}J$.
• B. $K=4.86\times {10}^{107};W_{max}=6.118\times {10}^{8}J$.
• C. $K=2.86\times {10}^{107};W_{max}=6.118\times {10}^{5}J$.
• D. none

Answer 1

The correct option is C $K=2.86\times {10}^{107};W_{max}=6.118\times {10}^{5}J$.

At equilibrium conditionwe have,

$E_{cell}^0$ = $\frac{0.059}{2}$ $logK$

0.80 + 2.37 = 0.0295 $logK$

$logK$ = 107.45

$K$ = 2.86 $\ast$ ${10}^{107}$

and $W_{max}$ = nF$E_{cell}^0$

$W_{max}$= 2$\ast$96500$\ast$3.17

$W_{max}$= 6.118 $\ast$ ${10}^5$ J