Question

What would be the electrode potential for the given half cell reaction at $pH =5$ ?
$2 H _{2} O \rightarrow O _{2}+4 H ^{\oplus}+4 e^{-} ;{E}_{\text {red }}^{0}=1.23 V$
$\left(R=8.314\, J\,mol ^{-1} \,K ^{-1} ; T_\text {emp}=298 K \right. \text {; }$ oxygen under std. atm. pressure of $1$ bar$)$
Given -

Answer 1

$2 H _{2} O (l) \rightarrow O _{2}(g)+4 H ^{+}+4 e^{-} ; E_{\text {red }}^{0}=1.23 V$
From nernst equation
$E_\text{cell}=E_\text{cell}^{0}-\frac{R T}{n F} \text{In} Q$
at $1$ bar & $298\, K$
$\frac{2.303 R T}{F}=0.059$
$p H=5 \Rightarrow\left[H^{+}\right]=10^{-5} M$
$E_{\text {oxidation }}^{\circ}=-1.23$ volt
$E_{\text {cell }}=-1.23-\frac{0.059}{4} \log \left(10^{-5}\right)^{4}$
$=-1.23+0.059 \times 5=-0.935\, V$