Question

What will be the effect on emf of cell when the concentration of $M{g}^{2+}$ ion is decreased to ${10}^{-6}M$ at ${25}^{o}C$?
$Mg\left(s\right)|M{g}^{2+}(aq,0.2M)||A{g}^{+}(aq,1\times {10}^{-3})|Ag\left(s\right)$
$E_{A{g}^{+}/Ag}^0=+0.8V$
$E_{M{g}^{2+}/Mg}^0=-2.37V$

Answer 1

For the given cell, standard reduction potential of $A{g}^{+}/Ag$ couple is higher than $M{g}^{2+}/Mg$ couple.
Hence, silver electrode will act as cathode and magnesium electrode will act as anode.

$E_{\text{cell}}^0=E_{\text{cathode}}^0-E_{\text{anode}}^0$
$E_{\text{cell}}^0=0.80-(-2.37)=3.17V$

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

By Nernst equation,

$E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{n}log\frac{\left[M{g}^{2+}\right]}{[A{g}^{+}{]}^2}$
$E_{\text{cell}}=3.17-\frac{0.0591}{2}log\frac{0.2}{[1\times {10}^{-3}{]}^2}$
$E_{\text{cell}}=3.17-\frac{0.0591}{2}log(2\times {10}^5)$
$E_{\text{cell}}=3.17-\frac{0.0591}{2}\times 0.301-\frac{0.0591}{2}\times 5$
$E_{\text{cell}}=3.17-0.156=3.014V$

When $\left[M{g}^{2+}\right]={10}^{-6}M$
$E_{\text{cell}}=E_{\text{cell}}^0-\frac{0.0591}{2}log\frac{{10}^{-6}}{(1\times {10}^{-3}{)}^2}$
$E_{\text{cell}}=3.17-\frac{0.0591}{2}log\left(1\right)$
$E_{\text{cell}}=3.17V$

The emf of the cell increased from $3.014V$ to $3.17V$