Find the emf of the following cell reaction, given E Cr 3 + / cr 2 + 0 = - 0 . 72   V and E Fe 2 + / Fe…

The given cell is:

$\mathrm{Cr}\left|{\mathrm{Cr}}^{3+}(0.1\mathrm{M})\Vert {\mathrm{Fe}}^{2+}(0.1\mathrm{M})\right|\mathrm{Fe}$

The half reactions can be written as:

$$\begin{gathered} \mathrm{Cr}\to {\mathrm{Cr}}^{3+} +3{\mathrm{e}}^{-1} {\mathrm{E}}^{°} = -0.72\mathrm{V} \\ {\mathrm{Fe}}^{2+}+2{\mathrm{e}}^{-1} \to \mathrm{Fe} {\mathrm{E}}^{°} = -0.42\mathrm{V} \end{gathered}$$

The full reaction of the operating cell is:

$3{\mathrm{Fe}}^{2+}+2\mathrm{Cr} \to 2{\mathrm{Cr}}^{3+}+3\mathrm{Fe}$

Now, emf of the cell reaction is 

$$\begin{gathered} {{\mathrm{E}}^{°}}_{\mathrm{cell}} ={{\mathrm{E}}^{°}}_{\mathrm{cathode}}-{{\mathrm{E}}^{°}}_{\mathrm{anode}} \\ {{\mathrm{E}}^{°}}_{\mathrm{cell}}=-0.42-(-0.72)= 0.30 \mathrm{V} \end{gathered}$$

Therefore,

${\mathrm{E}}_{\mathrm{cell}} = {{\mathrm{E}}^{°}}_{\mathrm{cell}}-\frac{\mathrm{nRT}}{\mathrm{F}}\mathrm{lnQ}$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{cell}} = {{\mathrm{E}}^{°}}_{\mathrm{cell}} - \frac{2.303\mathrm{nRT}}{\mathrm{F}}{\log }_{10}\frac{{\left[{\mathrm{Cr}}^{3+}\right]}^2}{{\left[{\mathrm{Fe}}^{2+}\right]}^3} \\ {\mathrm{E}}_{\mathrm{cell}} = 0.3 - \frac{0.059}{6}{\log }_{10}\frac{{\left[0.1\right]}^2}{{\left[0.1\right]}^3} \\ {\mathrm{E}}_{\mathrm{cell}} \approx 0.30 \mathrm{V} \end{gathered}$$