Question

Calculate $E_{cell}$ and $\Delta$G for the following at ${28}^o$C

$Mg\left(s\right)+S{n}^{2+}\left(0.025M\right)\to M{g}^{2+}\left(0.06M\right)+S{n}_{\left(s\right)}$

$E_{cell}^0=2.23$V

Is the reaction spontaneous?

Answer 1

Given: Cell reaction
$M{g}_{\left(s\right)}+S{n}_{\left(aq\right)}^{2+}\to M{g}_{\left(aq\right)}^{2+}+S{n}_{\left(s\right)}$

$E_{Cell}=2.23;\left[S{n}^{2+}\right]=0.025$M

$\left[M{g}^{2+}\right]=0.06M;n=2$

$E_{cell}=E_{cell}-\frac{0.0592}{2}lo{g}_{10}\frac{\left[M{g}^{2+}\right]}{\left[S{n}^{2+}\right]}$

$=2.23-\frac{0.0592}{2}lo{g}_{10}\frac{0.06}{0.025}$

$=2.23-0.296log24$

$=2.23-0.296\times 0.380=2.118$V

$\Delta G=-nF{E}_{cell}$

As, $[1F=96500C]$

$\Delta G=-2\times 96500\times 2.118$

$=408774$J

$=-408.8$kJ

Since $\Delta$G for a given cell reaction is negative, the given reaction is spontaneous.