MCQ
What would be the electrode potential for the given half cell reaction at $pH = 5$ ? ........... $\mathrm{V}$

$2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{O}_{2}+4 \mathrm{H}^{\oplus}+4 \mathrm{e}^{-} ; \mathrm{E}_{\mathrm{red}}^{0}=1.23 \mathrm{V}$

$(\mathrm{R}=8.314 \;\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ; \text { Temperature }=298 \;\mathrm{K} ;$ oxygen under std. atm. pressure of $1 \;bar$)

  • $1.52$
  • B
    $2.56$
  • C
    $0.36$
  • D
    $3.56$

Answer

Correct option: A.
$1.52$
a
$\mathrm{O}_{2}(\mathrm{g})+4 \mathrm{H}^{+}+4 \mathrm{e}^{-} \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) ; \mathrm{E}_{\mathrm{red.}}^{0}=1.23 \mathrm{V}$

From nernst equation

$\mathrm{E}_{\text {cell }}=\mathrm{E}_{\text {cell }}^{0}-\frac{\mathrm{RT}}{\mathrm{nF}} \ln \mathrm{Q}$

at $1\; bar$ and $298 \mathrm{K}$

$\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059$

$\mathrm{pH}=5 \Rightarrow\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{M}$

$\mathrm{E}_{\text {cell }}=1.23-\frac{0.059}{4} \log \left[\mathrm{H}^{+}\right]^{4}$

$\mathrm{E}_{\text {cell }}=1.23-\frac{0.059}{4} \log \left(10^{-5}\right)^{4}$

$=1.23+0.295=1.525 \mathrm{V}$

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