Question

The emf of the cell with the cell reaction given below is $0.28V$ at ${25}^{o}C$.
$Zn\left(s\right)+2H^{+}\left(aq\right)\to Z{n}^{2+}(aq,0.1M)+H_2(g,1atm)$
Calculate the $pH$ of the hydrogen electrode
Given:
$E_{Z{n}^{2+}/Zn}^0=-0.76V$
$E_{H^{+}/H_2}^0=0V$

Answer 1

Standard reduction potential of $Z{n}^{2+}/Zn$ couple is less than standard reduction potential of hydrogen electrode.
Thus, hydrogen electrode will act as cathode and zinc couple will act as anode.

Here, standard hydrogen electrode is acting as reference electrode and its standard reduction potential is zero.
Hence, the standard cell potential is
$E_{cell}^0=E_{cathode}^0-E_{anode}^0$
$E_{cell}^0=0-(-0.76)$
$E_{cell}^o=0.76V$

The given cell reaction is,
$Zn\left(s\right)+2H^{+}\left(aq\right)\to Z{n}^{2+}(aq,0.1M)+H_2(g,1atm)$

According to Nernst equation,
$E_{cell}=E_{cell}^o-\frac{0.0591}{2}log\frac{\left[Z{n}^{2+}\right]P_{H_2}}{{\left[H^{+}\right]}^2}$
$0.28=0.76-\frac{0.0591}{2}log\frac{\left(0.1\right)\times 1}{{\left[H^{+}\right]}^2}$
$log\frac{\left(0.1\right)\times 1}{{\left[H^{+}\right]}^2}=\frac{2\times 0.48}{0.0591}$

$log0.1-log{\left[H^{+}\right]}^2=16.24$

$-log{\left[H^{+}\right]}^2=16.24-log0.1$

$-2log\left[H^{+}\right]=16.24+1$

$2pH=16.24+1\because -log\left[H^{+}\right]=pH$

$pH=\frac{17.24}{2}$

$pH=8.62$