Question

If the solution of $CuS{O}_4$ in which copper rod is immersed is diluted to $10$ times , the reduction electode potential:
$E_{C{u}^{+2}/Cu}^{\circ }$ = $0.34V$

• A. increased by 0.03 V
• B. decreased by 0.03 V
• C. increased by 0.059 V
• D. decreased by 0.059 V

Answer 1

The correct option is B decreased by 0.03 V

$C{u}^{2+}+2e^{-}⟶Cu$

$E_{cell}=E_{cell}^o-\frac{0.0591}{2}\log \frac{1}{\left[C{u}^{2+}\right]}$

The new concentration of $C{u}^{2+}$ is $\frac{1}{10}$ of $\left[C{u}^{2+}\right]$

$\Rightarrow E_{cell}^1=E_{cell}^o-\frac{0.0591}{2}\log \frac{10}{\left[C{u}^{2+}\right]}$

$\Rightarrow E_{cell}^1=E_{cell}^o-\frac{0.0591}{2}\log \frac{1}{\left[C{u}^{2+}\right]}-\frac{0.0591}{2}\log 10$

$\Rightarrow \Delta E_{cell}=-\frac{0.0591}{2}\log 10$

$\Rightarrow E_{cell}^1=-\frac{0.059}{2}$

Thus the electrode potential decrease by $\frac{0.059V}{2}=0.03V$