First slide

Electrochemical cells

Question

The standard electrode potential for the following reaction is +1.33 V. The potential at pH = 2.0 cannot be
${Cr}_{2} O_{7}^{2-} (aq,1M) + {14 H}^{+} (aq) + {6e}^{-} \to {2Cr}^{3+} (aq, 1M) + {7H}_{2} O (l)$

Moderate

A

$+ 1 .820 V$
B

$+ 1 .990 V$
C

$+ 1 .608 V$
D

$+ 1 .0542 V$

Solution

Given, Standard electrode potential E°= + 1.33V
$Electrode potential, E = ?, P^{H} = 2$
$E = E^{o} - 0 .059 / n \log \frac{[OP]}{[RP]}$
$\Rightarrow E_{{Cr}_{2} O_{7}^{2-} / {Cr}^{3+}} = E_{{Cr}_{2} O_{7}^{2-} / {Cr}^{3+}}^{o} - \frac{0 .0591}{6} × (log \frac{{[{Cr}^{3+}]}^{2}}{[{Cr}^{2} O_{7}^{2-}]} × \frac{1}{{[H^{+}]}^{14}})$
$\Rightarrow E_{{Cr}_{2} O_{7}^{2-} / {Cr}^{3+}} = 1 .33 - \frac{0 .0591}{6} \log \frac{1}{{(0 .01)}^{14}}$
$\Rightarrow 1.0542 V$

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