Chemistry - Electrochemistry Question with Solution | TestHub

The reduction half-reaction for the copper electrode is:

${\text{Cu}}^{2+}+2e^{-}\to \text{Cu}$

The Nernst equation for the reduction potential is given by:

$E_{red}=E_{red}^0-\frac{0.05916}{2}\log \frac{1}{[{\text{Cu}}^{2+}]}$

Assuming the initial solution is in its standard state, $[{\text{Cu}}^{2+}]=1\text{ M}$, so $E_{red}=E_{red}^0$.

When the solution is diluted $10$ times, the new concentration becomes $[{\text{Cu}}^{2+}]=0.1\text{ M}$.

Substituting this into the Nernst equation:

$E_{red}=E_{red}^0-\frac{0.05916}{2}\log \frac{1}{0.1}$

$E_{red}=E_{red}^0-0.02958\log 10$

$E_{red}=E_{red}^0-0.02958\text{ V}$

Thus, the reduction potential is less by $0.02958\text{ V}$ than $E_{red}^0$.

Answer: Less by $0.02958$ V, than $E_{red}^0$