Question

For a silver- magnesium voltaic cell,
calculate the maximum useful work that can be obtained by operating the cell.
$M{g}^{2+}+2e^{-}\to Mg\left(s\right);E^0=-2.37V$
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right);E^0=+0.80V$

• A. $1002kJ$
• B. $815kJ$
• C. $612kJ$
• D. $334kJ$

Answer 1

The correct option is C $612kJ$

The standard reduction potential of silver is higher than magnesium.
Thus,
Cathode reaction is
$A{g}^{+}\left(aq\right)+e^{-}\to Ag\left(s\right)$

Anode reaction is
$Mg\left(s\right)\to M{g}^{2+}+2e^{-}$

$E_{cell}^0=\text{SRP of substance reduced}-\text{SRP of substance oxidised}$
(SRP - Standard reduction potential)

$E_{cell}^0=E_{cathode\left(red\right)}^0-E_{anode\left(red\right)}^0$

$E_{cell}^0=0.80+2.37=3.17V$

Cell reaction is
$2Ag\left(s\right)+M{g}^{2+}\left(aq\right)\to Mg\left(s\right)+2A{g}^{+}\left(aq\right)$
2 electrons are involved in the reaction.
Thus, $n=2$
$\text{Maximum useful work}=-\Delta G^0$
$\text{Maximum useful work}=nF{E}^0$
$\text{Maximum useful work}=2\times 96500\times 3.17=611810J$
$\text{Maximum useful work}=611.810kJ\approx 612kJ$