Question

For the cell reaction at ${25}^{\circ }C$,
$Mg\left(s\right)+2A{g}^{+}\left(aq\right)\rightleftharpoons M{g}^{2+}\left(aq\right)+2Ag\left(s\right)$
$E_{cell}^0=+3.17V$
The values of $E_{cell},\Delta G^0\text{and}Q$ respectively , at $\left[A{g}^{+}\right]={10}^{-3}M$ and $\left[M{g}^{2+}\right]=0.02M$ are:
Here,
$Q\text{is reaction quotient}$
$log2=0.301$

• A. $3.30V,-605.8kJmo{l}^{-1},2000$
• B. $3.04V,-586.7kJmo{l}^{-1},2000$
• C. $3.13V,-604kJmo{l}^{-1},20$
• D. $3.04V,-611.8kJmo{l}^{-1},20000$

Answer 1

The correct option is D $3.04V,-611.8kJmo{l}^{-1},20000$

For a general electrochemical reaction,
$aA+bB\leftrightharpoons cC+dD$

Nernst equation is,
$E=E^0-\frac{2.303RT}{nF}log\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}.....(Eqn.1)$
$Q=\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}$
Q is reaction quotient
n is number of electron involved in the reaction

Substituting,

$R=8.314J{K}^{-1}mo{l}^{-1}$

$F=96500C/mol$

$T=25+273=298K$, we get,

$E_{cell}=E_{cell}^0-\frac{0.059}{n}log\frac{\left[C{]}^c\right[D{]}^d}{\left[A{]}^a\right[B]b}$

For the cell reaction,
$Mg\left(s\right)+2A{g}^{+}\left(aq\right)\rightleftharpoons M{g}^{2+}\left(aq\right)+2Ag\left(s\right)$

$E^0=+3.17V,n=2$
$Q=\frac{\left[M{g}^{2+}\right]}{[A{g}^{+}{]}^2}=\frac{0.02}{[0.001{]}^2}=20000$


$E_{cell}=E_{cell}^0-\frac{0.059}{n}logQ$
$E_{cell}=3.17-\frac{0.059}{2}log\left(20000\right)$
$E_{cell}=3.17-\frac{0.059}{2}log(2\times {10}^4)$
$E_{cell}=3.17-\frac{0.059}{2}\times 0.301-\frac{4\times 0.059}{2}$
$E_{cell}=+3.04V$

$\Delta G^0=-nF{E}^0$
$\Delta G^0=-2\times 96500\times 3.17$
$\Delta G^0=-611.8kJmo{l}^{-1}$