Calculate the electrode potential at 250C of Cr2O7−2/Cr+3electrode at POH=11 in a solution of 0.01M both in Cr+3and Cr2O7−2, EMF of the cell is Cr2O7−2+14H⊕+6e−→2Cr+3+7H2O, EO=1.33V

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Calculate the electrode potential at $25^{0} C$ of $C r_{2} {O_{7}}^{- 2} / C r^{+ 3}$ electrode at $P^{O H} = 11$ in a solution of $0.01$ M both in $C r^{+ 3}$ and $C r_{2} {O_{7}}^{- 2}$ , EMF of the cell is $C r_{2} {O_{7}}^{- 2} + 14 H^{\oplus} + 6 e^{-} \to 2 C r^{+ 3} + 7 H_{2} O, E^{O} = 1.33 V$

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$0.652 V$

$0.213 V$

$0.725 V$

$0.936 V$

answer is B.

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Detailed Solution

$C r_{2} {O_{7}}^{- 2} + 14 H^{\oplus} + 6 e^{-} \to 2 C r^{+ 3} + 7 H_{2} O, E^{O} = 1.33 v$
$0.01 10^{- 3} 0.01$
$P^{O H} = 11$
$P^{H} = 3$
$[H^{+}] = 10^{- 3} M$
E cell $= E^{0} c e l l - \frac{0.0591}{6} \log \frac{{[c r^{+ 3}]}^{2}}{[c r_{2} {O_{7}}^{- 2}] {[H^{+}]}^{14}}$
$= 1.33 - \frac{0.0591}{6} \log \frac{{[0.01]}^{2}}{(0.01) {(10^{- 3})}^{14}} = 0.936 V$