Q.13E - Chemistry (OpenStax)

Question

Find the potentials of the following electrochemical cell:

\(Cd|C{d^{2 + }},M = 0.10||N{i^{2 + }},M = 0.50|Ni\)

Step-by-Step Solution

Answer

The potentials of the following electrochemical cell is \({E_{cell}} = 0.167V\)

Step 1: solving using Nernst equation:

We can use Nernst equation; 

\({E_{cell}} = {E_{cell}} - \frac{{0.05915}}{n}.\log \frac{{product}}{{reac\tan t}}\)

 

Standard electrical potential,

\(C{d^{2 + }} + 2{e^ - } \to Cd,\)    E = -0.403V

\(N{i^{2 + }} + 2{e^ - } \to Ni,\)      E = -0.257V

Step 2: Calculating the standard electrode potential:

Calculate the standard electrode potential,

\(\begin{aligned}{l}{E_{cell}} &= {E_{cathode}} - {E_{anode}}\\{E_{cell}} &= \left( { - 0.257V} \right) - \left( { - 0.403V} \right)\\{E_{cell}} &=   - 0.146V\end{aligned}\)

 

We have reaction,

\(Cd + N{i^{2 + }}\left( {0.5M} \right) \to Ni + C{d^{2 + }}(0.1M)\)

We can see that \(N{i^{2 + }}\)is our reactant and \(C{d^{2 + }}\)is our product,

\(\begin{aligned}{l}{E_{cell}} &= {E_{cell}} - \frac{{0.05915}}{n}.\log \frac{{product}}{{reac\tan t}}\\{E_{cell}} &=  - 0.146V - \frac{{0.05915}}{2}.\log \frac{{(0.1M)}}{{(0.5M)}}\\{E_{cell}} &= 0.167V\end{aligned}\)