Question

A galvanic cell is constructed of two hydrogen electrodes, one immersed in a solution with $H^{+}$ at $1M$ and the other in $1M$ $KOH$. Calculate $E_{cell}$. If $1M$ $KOH$ solution is replaced by $1M$ ${NH}_3$, will $E_{cell}$ be higher or lower than in $1M$ $KOH$?

Answer 1

$\left[H^{+}\right]$ in one cell $=1M$

$\left[{OH}^{-}\right]$ in another half cell $=1M$

$\left[H^{+}\right]\left[{OH}^{-}\right]={10}^{-14}$

$\left[H^{+}\right].1={10}^{-14}\Rightarrow \left[H^{+}\right]={10}^{-14}M$

In concentration cell, half cell with less $\left[H^{+}\right]$ act as anode

$\frac{1}{2}{H}_2\to H_{Anode}^{+}+e^{-}$

$\underset{–}{H_{cathode}^{+}+e^{-}\to \frac{1}{2}{H}_2}$
$H_{cathode}^{+}\to H_{anode}^{+}$

$Q=\frac{\left[H^{+}\right]anode}{\left[H^{+}\right]cathode}=\frac{{10}^{-14}}{1}$

$\therefore {\epsilon }_{cell}=\frac{-0.0592}{1}\log {10}^{-14}=14\times 0.0592=0.83V$

$1M$ ${NH}_3$ has lower ${OH}^{-}$ concentration because it is a weaker base. So $\left[H^{+}\right]$ will be higher

So, $\log \left[H^{+}\right]$ will be higher and hence $-0.0592\log \left[H^{+}\right]$ will be lower