Electrochemistry Solved Problems

Here are a few simple electrochemistry solved problems along with their solutions:

1. Calculate the single electrode potential of copper metal immersed in $0.15°MCu 2+$ solution. $E°$ for copper is $+0.34 V$ at $25°C$ .

Solution:

By Nernst equation $\mathrm{E}=\mathrm{E}^{\circ}+\frac{0.0592}{\mathrm{n}}\left[\log \mathrm{M}^{\mathrm{n}+}\right] at 25^{\circ} \mathrm{C}$

$\begin{aligned} & =0.34+\frac{0.0592}{2} \log [0.15] \\ & =0.34+\frac{0.0592}{2}(-0.8239)=0.3156 \mathrm{~V} \end{aligned}$

2. Calculate the electrode potential of lead immersed in its metal ion solution of $0.015M$ concentration. The standard electrode potential for the reaction

Pb → Pb²⁺ + 2e^– is 0.13V.

Solution:

The reaction $Pb⟶Pb 2+ +2e -$ is an oxidation reaction and the corresponding potential $0.13°V$ is standard oxidation potential $(E pb/$ $Pb 2+$ ). During substitution, reduction potential should be substituted in the formula of Nernst equation.

So, the reduction reaction is

Pb2⁺ + 2e^– → Pb and

E°_Reduction (i.e.,) E_Pb²⁺ / _Pb = -0.13°V

By Nernst equation $\mathrm{E}=\mathrm{E}^{\circ}+\frac{0.0592}{2} \log \left[\mathrm{M}^{\mathrm{n}+}\right] at 25^{\circ} \mathrm{C}$

$\begin{aligned} \mathrm{E} & =-0.13+\frac{0.0592}{2} \log [0.015] \end{aligned}$

= -0.13 + 0.02955 (-1.824)

= -0.1828°V.

3. Find the potential of the cell in which the following reactions take place at $25°C$ .

Zn(s) + Cu²⁺ (0.02°M) → Cu(s) + Zn²⁺ (0.4M)

Given E° (Zn²⁺ / Zn) = -0.76°V

E°(Cu²⁺ / Cu) = 0.34°V

Solution:

The standard reduction potential $E°$ of copper (0.34) is higher than $E°$ of zinc $(-0.76 V)$ . So zinc behaves as anode and copper as cathode. So the cell can be represented as,

Zn | Zn²⁺ (0.4M) || Cu²⁺ (0.02M) | Cu

E_cell = E_cathode -E_anode

$\begin{aligned} & \quad=\mathrm{E}_{\mathrm{Cu}}^{\circ}+-\frac{0.0592}{2} \log \left[\mathrm{Cu}^{2+}\right]-\mathrm{E}_{\mathrm{Zn}}^{\circ}+-\frac{0.0592}{2} \log \left[\mathrm{Zn}^{2+}\right] \end{aligned}$

$\begin{aligned} & =\left[\mathrm{E}_{\mathrm{Cu}}^{\circ}-\mathrm{E}_{\mathrm{Zn}}\right]+\frac{0.0592}{2} \log \frac{\left[\mathrm{Cu}^{2+}\right]}{\left[\mathrm{Zn}^{2+}\right]} \\ & =[0.34-(-0.76)]+-\frac{0.0592}{} \frac{0.02}{2} \frac{0.04}{} \\ & =1.1+\frac{0.0592}{2} \log (0.05) \\ \end{aligned}$