Question

For the cell reaction,
$Mg ( s )+2 Ag ^{+}( aq ) \rightleftharpoons Mg ^{2+}( aq )+2 Ag ( s )$
$E _{\text {cell }}^{ o }$ is $3.17 \,V$ at $298 \,K$. The value of $E _{\text {cell }}, \Delta G ^{\circ}$ and $Q$ at $Ag ^{+}$and $Mg ^{2+}$ concentrations of $0.001$ $M$ and $0.02\, M$ respectively are :

• A. $3.04\, V ,-605.8 \,kJ \,mol ^{-1}, 20000$
• B. $3.04 \,V , 611.8\, kJ \,mol ^{-1}, 20000$
• C. $3.13 \,V ,-604\, kJ \,mol ^{-1}, 20$
• D. $3.04\, V ,-611.8 \,kJ , 20000$

Answer 1

Answer: (D)

$ Mg ( s )+2 Ag ^{+}( aq ) \rightleftharpoons Mg ^{+2}( aq )+2 Ag ( s ) $
$ E _{ cell }= E _{\text {cell }}^{\circ}-\frac{0.0591}{ n } \log \frac{\left[ Mg ^{+2}\right]}{\left[ Ag ^{+}\right]^{2}}$
$ E _{ cell }=3.17-\frac{0.0591}{2} \log \frac{\left(2 \times 10^{-2}\right)}{\left(10^{-3}\right)^{2}} $
$E _{\text {cell }}=3.17-\frac{0.0591}{2} \times 4.3 $
$=3.17-0.129 $
$=3.04 \,volt $
$\Delta G ^{\circ} =- nF E _{\text {cell }}^{\circ} $
$=-2 \times 96500 \times 3.17 $
$=-611.8\, kJ $
$Q =\frac{\left[ Mg ^{2+}\right]}{\left[ Ag ^{+}\right]^{2}}=20000$
$Q =20000$