Question

For the cell $Mg\left(s\right)|M{g}^{2+}\left(aq\right)||A{g}^{+}\left(aq\right)|Ag\left(s\right)$, calculate the equilibrium constant at ${25}^o$C and the maximum work that can be obtained during operation of cell.

Given $E_{Mg/M{g}^{2+}}^o=+2.37V$ and$E_{A{g}^{+}/Ag}^o=0.80V$

• A. $6.118\times {10}^{2}kJ$
• B. $4.315\times {10}^{2}kJ$
• C. $2.128\times {10}^{2}kJ$
• D. None of these

Answer 1

The correct option is A $6.118\times {10}^{2}kJ$

$E_{cell}=\left(0.059/2\right)log{K}_c$

$\because E_{cell}=E_{O{P}_{Mg}}^o+E_{R{P}_{Ag}}^o$ for the change

$Mg+2A{g}^{+}\to M{g}^{2+}+2Ag$

or $E_{O{P}_{Mg}}^o+E_{R{P}_{Ag}}^o=\left(0.059/2\right)log{K}_c$

or $2.37+0.80=\left(0.059/2\right)log{K}_c$

or $log{K}_c=107.457$ and $E_{cell}^o=3.17V$

For maximum work $-\Delta G^o=W_{max}$

$W_{max}=nF\times E^o=2\times 96500\times 3.17=6.118\times {10}^{5}joule$

$=6.118\times {10}^{2}kJ$

Hence, the correct option is $\text{A}$