Question

The change in potential of the half-cell $Cu^{2+}|Cu$, when aqueous $Cu^{2+}$ solution is diluted $100$ times at $298 \,K$ ? $\left( \frac{2.303 RT}{F} = 0.06 \right)$

• A. increases by 120 mV
• B. decreases by 120 mV
• C. increases by 60 mV
• D. decreases by 60 mV
• E. no change

Answer 1

Answer: (D)

Half cell reaction is

$Cu ^{2+}+2 e^{-} \longrightarrow Cu ; n=2$

From Nernst equation,

$E =E^{\circ}-\frac{2.303\, R T}{n F} \log \frac{1}{\left[ Cu ^{2+}\right]} $

$E_{1} =E^{\circ}+\frac{0.06}{2} \log \left[ Cu ^{2+}\right]$

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

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

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

$\therefore \underset{\text{Initial}}{M_{1} V_{1}}=\underset{\text{After dilution}}{M_{2} V_{2}}$

$ 1 \times 1 =M_{2} \times 100 $

$M_{2} =\frac{1}{100}=0.01 $

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

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

$=F_{1}-0.059\, V =F_{1}-59\, mV$

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