Question

Write the Nernst equation and emf of the following cells at 298 K:

(i) $Mg\left(s\right)|M{g}^{2+}\left(0.001M\right)||C{u}^{2+}\left(0.001M\right)|Cu\left(s\right)$

(ii) $Fe\left(s\right)|F{e}^{2+}\left(0.001M\right)||H^{+}\left(1M\right)|H_2\left(g\right)\left(1bar\right)|Pt\left(s\right)$

(iii) $Sn\left(s\right)|S{n}^{2+}\left(0.050M\right)||H^{+}\left(0.020M\right)|H_2\left(g\right)\left(1bar\right)|Pt\left(s\right)$

(iv) $Pt\left(s\right)|B{r}_2\left(l\right)|B{r}^{-}\left(0.010M\right)||H^{+}\left(0.030M\right)|H_2\left(g\right)\left(1bar|Pt\right(s\left)\right)$

$E_{\left(\frac{M{g}^{2+}}{Mg}\right)}^{\circ }=-2.37V;E_{\left(\frac{C{u}^{2+}}{Cu}\right)}^{\circ }=+0.34V;E_{\left(\frac{F{e}^{2+}}{Fe}\right)}^{\circ }=-0.44V;E_{\left(\frac{S{n}^{2+}}{Sn}\right)}^{\circ }=-0.14V;E_{\left(\frac{\frac{1}{B{r}_2}}{B{r}^{-}}\right)}^{\circ }=+1.08V,$

Answer 1

Nernst equation: It is a mathematical equation that is used to calculate the emf of a cell.

Cell representation:

• The left-hand side of a cell represents the anode where the oxidation takes place.
• The right-hand side of a cell represents the cathode where reduction takes place.
• The middle part of a cell representation denotes the salt bridge.

(i)

$Cellequation:Mg\left(s\right)+C{u}^{2+}\left(aq\right)\to M{g}^{2+}\left(aq\right)+Cu\left(s\right)$

$(n=2)$

Nernst equation : $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{\left[M{g}^{2+}\right]}{C{u}^{2+}}$

EMF of the cell:

$E_{cell}=[0.34-(-2.37\left)\right]-\frac{0.0591}{2}log\frac{\left[{10}^{-3}\right]}{{10}^{-4}}$

$=2.71-0.02955=2.68V$

$EMF=2.68V$

(ii)

Cell equation : $Fe\left(s\right)+2H^{+}\left(aq\right)\to F{e}^{2+}\left(aq\right)+H_2\left(g\right)$

$(n=2)$

Nernst equation : $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{\left[F{e}^{2+}\right]}{[H^{+}{]}^2}$

EMFof the cell,

$E_{cell}=[0-(-0.44\left)\right]\frac{0.0591}{2}log\frac{\left[{10}^{-3}\right]}{[1{]}^2}$

$=0.44-\frac{0.591}{2}\times (-3)$

$=0.44+0.0887$

$=0.5287V\approx 0.53V$

$EMF=0.53V$

(iii)

$CellEquation:Sn\left(s\right)+2H^{+}\left(aq\right)\to S{n}^{2+}\left(aq\right)+H_2\left(g\right)$

$(n=2)$

Nernst equation ; $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{S{n}^{2+}}{[H^{+}{]}^2}$

EMF of the cell,

$E_{cell}=[0-(-0.14)-\frac{0.0591}{2}log\frac{\left[0.05\right]}{[0.02{]}^2}]$

$=0.14-\frac{0.0591}{2}\times \left(2.097\right)$

= 0.14 - 0.0620 = 0.08V

$EMF=0.08V$

(iv)

Cell equation : $2B{r}^{-}\left(l\right)+2H^{+}\left(aq\right)\to B{r}_2\left(l\right)+H_2\left(g\right)$

$(n=2)$

Nernst equation : $E_{cell}=E_{cell}^{\circ }-\frac{0.0591}{2}log\frac{1}{\left[B{r}^{-}{]}^2\right[H^{+}{]}^2}$

EMF of the cell,

$E_{cell}=[0-1.08]-\frac{0.0591}{2}log\frac{1}{(0.01{)}^2\times (0.03{)}^2}$

$=-1.08-\frac{0.0591}{2}log(\mathrm{1.11.1}\times {10}^7)$

$=-1.08-\frac{0.0591}{2}\left(7.0457\right)$

$=-1.08-0.208=-1.288V$

$EMF=-1.288V$

($\therefore$ Oxidation will occur at hydrogen electrode and reduction at $B{r}_2$ electrode)