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°MCu2+ 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⟶Pb2++2e− is an oxidation reaction and the corresponding potential 0.13°V is standard oxidation potential (Epb/ Pb2+ ). 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.,) EPb2+ / 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
Ecell = Ecathode -Eanode
\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}= 1.1 + (-0.0385) = 1.0615V.
E°cell = 0.34 – (-0.44)
=0.78V.
| Read More Topics |
| Measurement of single electrode potential |
| Definition and origin of electrode potential |
| Crystal physics – Solved Problems |
Above are a couple of electrochemistry solved problems
