Question

By how much will the potential of half cell $C{u}^{2+}\left(0.1\text{M}\right)/Cu$ change, If the solution is diluted to $100$ times at $298\text{K}$?

• A. Increase by $59\text{mV}$
• B. Decreases by $\text{59 mV}$
• C. Increased by $\text{29.5 mV}$
• D. Decreases by $29.5\text{mV}$

Answer 1

The correct option is D Decreases by $29.5\text{mV}$

The half-cell reaction is $C{u}^{2+}+2e^{-}\to Cu$
So, the number of electrons involved is $2$.
$\therefore n=2$
$E_{\text{cell}}=E_{\text{Cell}}^{\circ }-\frac{0.059}{n}\text{log}\frac{1}{\left[C{u}^{2+}\right]}\Rightarrow E_{\text{cell}}=E_{\text{Cell}}^{\circ }-\frac{0.059}{2}\text{log}\frac{1}{\left[C{u}^{2+}\right]}$
According to the question after dilution the concentration of $C{u}^{2+}$ becomes $\frac{\left[C{u}^{2+}\right]}{100}$.
$\therefore E_{\text{cell}}^{\prime }=E_{\text{cell}}^{\circ }-\frac{0.059}{n}\text{log}\frac{1}{\left[\frac{C{u}^{2+}}{100}\right]}$
$\Rightarrow E_{\text{cell}}^{\prime }=E_{\text{cell}}^{\circ }-\frac{0.059}{2}\text{log}\frac{{10}^2}{\left[C{u}^{2+}\right]}$
$\Rightarrow E_{\text{cell}}^{\prime }=E_{\text{cell}}^{\circ }-\frac{0.059}{2}\times \text{2 log}\left(\frac{1}{\left[C{u}^{2+}\right]}\right)$
$\Rightarrow E_{\text{cell}}^{\prime }=E_{\text{cell}}^{\circ }-0.059\text{log}\frac{1}{\left[C{u}^{2+}\right]}$
So, the decrease in potential is
$E_{\text{cell}}^{\prime }-E_{\text{cell}}=-0.059\text{log}\frac{1}{\left[C{u}^{2+}\right]}+\frac{0.059}{2}\text{log}\frac{1}{\left[C{u}^{2+}\right]}=-\frac{0.059}{2}\text{log}\frac{1}{\left[C{u}^{2+}\right]}=-\frac{0.059}{2}\text{log}\frac{1}{{10}^{-1}}=-0.0295\text{V}or-29.5\text{mV}$

Hence Potential of half-cell will decrease by $29.5\text{mV}$.