Question

What would be the electrode potential for the given half cell reaction at $pH=5$ ?
$2H_{2}O\to O_2+4H^{\oplus }+4e^{-};E_{\text{red }}^0=1.23V$
$(R=8.314Jmo{l}^{-1}{K}^{-1};T_{\text{emp}}=298K\text{; }$ oxygen under std. atm. pressure of $1$ bar$)$
Given -

Answer 1

Answer: 1.52

$2H_{2}O(l)\to O_2(g)+4H^{+}+4e^{-};E_{\text{red }}^0=1.23V$
From nernst equation
$E_{\text{cell}}=E_{\text{cell}}^0-\frac{RT}{nF}\text{In}Q$
at $1$ bar & $298K$
$\frac{2.303RT}{F}=0.059$
$pH=5\Rightarrow \left[H^{+}\right]=10^{-5}M$
$E_{\text{oxidation }}^{\circ }=-1.23$ volt
$E_{\text{cell }}=-1.23-\frac{0.059}{4}\log {\left(10^{-5}\right)}^4$
$=-1.23+0.059\times 5=-0.935V$