Question

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

• 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

$C{u}^{2+}+2e^{-}⟶Cu;n=2$

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 dilued to $100$ times.

$\therefore \underset{\text{Initial}}{M_1{V}_1}=\underset{\text{After dilution}}{M_2{V}_2}$

$1\times 1=M_2\times 100$

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

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

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

$=F_1-0.059V=F_1-59mV$

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