Question:
What would be the electrode potential for the given half cell reaction at $\mathrm{pH}=5$ ? ______________.
$2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{\oplus}+4 \mathrm{e}^{-} ; E_{\mathrm{red}}^{0}=1.23 \mathrm{~V}$
$\left(\mathrm{R}=8.314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ;\right.$ Temp $=298 \mathrm{~K} ;$ oxygen under std. atm. pressure of 1 bar)
Solution:
(1.52)
$\mathrm{E}=1.23-\frac{0.0591}{4} \log \left[\mathrm{H}^{+}\right]^{4}$
$=1.23+0.0591 \times \mathrm{pH}$
$=1.23+0.0591 \times 5$
$=1.23+0.2955$
$=1.52 \mathrm{~V}$
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