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

Answer 1

The Correct Option is 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 $59(\approx 60) \,mV$.

Answer 2

Given:
Half-cell reaction: \(\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \ , \ n = 2\)
Nernst equation: \(E = E^\circ - \frac{{2.303RT}}{nF} \log \frac{1}{{[\text{Cu}^{2+}]}}\)

\(\frac{{2.303RT}}{nF} = 0.06\)

Initial concentration of \(\text{Cu}^{2+}: [\text{Cu}^{2+}] = 1  M\)
After dilution, the concentration becomes \(0.01\)M
We need to find the change in potential, \(\Delta E = E_1 - E_2\)

Let's calculate \(E_1\), the initial potential:

\(E_1 = E^\circ + \frac{{0.06}}{2} \log \left( \left[ \text{Cu}^{2+} \right] \right)\)

Since the initial concentration is 1M:

\(E_1 = E^\circ + \frac{{0.06}}{2} \log (1)\)
\(E_1 = E^\circ + 0\)
\(E_1 = E^\circ\)

Now, after dilution, the concentration becomes 0.01M. So, \(E_2\) can be calculated as:
\(E_2 = E^\circ + \frac{{0.059}}{2} \log \left( 0.01 \right)\)

\(E_2 = E^\circ + \frac{{0.059}}{2} (-2)\)

\(E_2 = E^\circ - 0.059 \, \text{V}\)
So, the potential decreases by \(59 \, \text{mV}\). This is approximately \(60 \, \text{mV}\).

So, the correct option is (D): decreases by 60 mV