Question

If hydrogen electrode is dipped in $2$ solutions of $pH =4$ and $pH =6$ at $1\, atm$ pressure and salt bridge is connected; the e.m.f. of resulting cell is_______ $V$.

Answer 1

$pH =4,\left[ H ^{+}\right]=10^{-4} M$
$pH =6,\left[ H ^{+}\right]=10^{-6} M$
The, electrode which has lower $\left[ H ^{+}\right]$acts as anode while electrode with higher $\left[ H ^{+}\right]$acts as cathode.
Hence,
Anode reaction:
$H _{2}\left( P _{1}\right) \longrightarrow 2 H ^{+}\left(10^{-6} M \right)+2 e ^{-}$
Cathode reaction:
$2 H ^{+}\left(10^{-4} M \right)+2 e ^{-} \longrightarrow H _{2}\left( P _{2}\right)$
$E _{\text {anode }}=-\frac{ RT }{2 F } \ln \frac{\left[ H ^{+}\right]^{2}}{ P _{1}}=-\frac{0.0591}{2} \log \frac{\left(10^{-6}\right)^{2}}{1}$
$E _{\text {cathode }}=-\frac{ RT }{2 F } \ln \frac{ P _{2}}{\left[ H ^{+}\right]^{2}}-\frac{0.0591}{2} \log \frac{1}{\left(10^{-4}\right)^{2}}$
$E _{\text {cell }}= E _{\text {anode }}+ E _{\text {cathode }}$
$=-\frac{0.0591}{2} \log \frac{\left(10^{-6}\right)^{2}}{1}-\frac{0.0591}{2} \log \frac{1}{\left(10^{-4}\right)^{2}}$
$E _{\text {cell }}=-\frac{0.0591}{2} \log \frac{\left(10^{-6}\right)^{2}}{\left(10^{-4}\right)^{2}}$
$=-0.0591 \log \frac{10^{-6}}{10^{-4}}$
$=-0.0591 \times \log 10^{-2}$
$=-0.0591 \times(-2)=0.12\, V$