Question

The change in potential of the half-cell $C{u}^{2+}/Cu$, when aqueous $C{u}^{2+}$ solution is diluted $100$ times at $298K?$

$[\frac{2.303RT}{F}=0.06]$.

• A. Increases by $120mV$
• B. Decreases by $120mV$
• C. Increases by $60mV$
• D. Decreases by $60mV$
• E. No change

Answer 1

The correct option is D Decreases by $60mV$

Half cell reaction is
$C{u}^{2+}+2e^{-}\to Cu;n=2$

From Nernst equation,
$E=E^{\circ }-\frac{2.303RT}{nF}\log \frac{1}{\left[C{u}^{2+}\right]}$

$E_1=E^{\circ }+\frac{0.06}{2}\log \left[C{u}^{2+}\right]$

Let the initial concentration of $C{u}^{2+}$ be $1$.

$E_1=E^{\circ }+\frac{0.06}{2}\log 1=E^{\circ }+0\therefore E_1=E^{\circ }$

Further, the $\left[C{u}^{2+}\right]$ solution is diluted to $100times$.

$\therefore \underset{Initial}{M_1{V}_1}=\underset{Afterdilution}{M_2{V}_2}$

$1\times 1=M_2\times 100$

$M_2=\frac{1}{100}=0.01$

$\therefore E_2=E^{\circ }+\frac{0.059}{2}\log \left[0.01\right]$

$=E^{\circ }+\frac{0.059}{2}(-2)$

$=E_1-0.059V=E_1-59mV$

Thus, the potential decreases by $59(\approx 60)mV$.