Question

By how much will the potential of half cell $C{u}^{2+}/Cu$ change, if the solution is diluted to 100 times at $298K$?

• A. Increases by 59.10 $mV$
• B. Decreases by 59.10 $mV$
• C. Increases by 29.55 $mV$
• D. Decreases by 29.55 $mV$

Answer 1

The correct option is B Decreases by 59.10 $mV$

For $C{u}^{2+}+2e^{-}\to Cu\left(s\right)$
applying nearst equation:
$E_{C{u}^{2+}/Cu}=E^0-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$

When $C{u}^{2+}$ solution is diluted to 100 times $\left[C{u}^{2+}\right]$ decreases to 1/100
Now, let the new potential after dilution be $E_{C{u}^{2+}/Cu}^{\prime }$
So,
$E_{C{u}^{2+}/Cu}^{\prime }=E^0-\frac{0.0591}{2}log\frac{100}{\left[C{u}^{2+}\right]}$
$E_{C{u}^{2+}/Cu}^{\prime }=E^0-\frac{0.0591}{2}[log100-log\left[C{u}^{2+}\right]]$
$\therefore$$E_{C{u}^{2+}/Cu}^{\prime }=E^0-\frac{0.0591}{2}\times 2-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$
$\therefore$$E_{C{u}^{2+}/Cu}^{\prime }=E^0-0.0591-\frac{0.0591}{2}log\frac{1}{\left[C{u}^{2+}\right]}$
$\therefore$$E_{C{u}^{2+}/Cu}^{\prime }=E_{C{u}^{2+}/Cu}-0.0591$
hence, the potential is decreased by $59.1mV$