First slide

Nernst Equation

Question

A hydrogen gas electrode is made by dipping platinum wire in a solution of HCI and pH = 10 and by passing hydrogen gas around the platinum wire at 1 atm pressure. The oxidation potential of electrode would be

Easy

A

0.059 V
B

0.59 V
C

0.118 V
D

0.18 V

Solution

For hydrogen electrode, oxidation half-reaction is $\underset{(1 atm)}{H_{2}} ⟶ \underset{at {pH}_{10}}{2 H^{+}} + 2 e^{-}$
$If pH = 10, H^{+} = 1 \times 10^{- pH} = 1 \times 10^{- 10}$
From Nernst equation, $E = E^{\circ} - \frac{0 .0591}{2} \log \frac{{[H^{+}]}^{2}}{p_{H_{2}}}$
For hydrogen electrode, $E^{\circ} = 0$
$\begin{array}{l} E = - \frac{0 .0591}{2} \log \frac{{(10^{- 10})}^{2}}{1} = - \frac{0 .0591 \times 2}{2} \log (10)^{- 10} \\ = 0 .0591 \times 10 \times \log 10 = 0 .59 V \end{array}$

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