The oxidation potential of a hydrogen electrode at $p H = 10$ and $P_{H_{2}} = 1$ :-

0.51 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.00 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

+ 0.59 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.059 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $+ 0.59 V$
For hydrogen electrode : $2 H^{+} + 2 e^{-} \to H_{2}$
$E_{r e d} = E_{r e d}^{o} - \frac{0.0592}{n} l o g \frac{P_{H 2}}{[H^{+}]^{2}}$
$E_{r e d} = 0 - \frac{0.0592}{2} l o g \frac{1}{(10^{- 10})^{2}}$
$E_{r e d} - 0.59 V$
$E_{o x i d} = - E_{r e d} = - (- 0.59) = 0.59 V$
Note: $[H^{+}] = 10^{- p H} = 10^{- 10}$

Suggest Corrections

Similar questions

The Oxidation potential of hydrogen electrode at $P^{H} = 10$ and $P_{H_{2}}$ at 1atm is :

_{H_{2}}

p H = 1

T = 298 K

p H

0.059 V

Join BYJU'S Learning Program