A hydrogen gas electrode is made by dipping platinum wire in a solution of $H C l$ of $p H$ = 10 and bypassing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of the electrode would be?

0.059 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

0.118 V

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

1.18 V

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

Open in App

Solution

The correct option is C $0.59 V$
$p H = 10 p H = - log [H^{+}] [H^{+}] = 10^{- 10} \frac{1}{2} H_{2} \to H^{+} + e^{-} E = E^{0} - \frac{0.0591}{1} l o g \frac{[H^{+}]}{f_{H_{2}}^{1 / 2}} f_{H_{2}}^{1 / 2} = P_{H_{2}}^{1 / 2} = ({1)}^{1 / 2} = 1 E = 0 - \frac{0.0591}{1} l o g 10^{- 10} = 0.59 V$

Option B is correct.

Suggest Corrections

Similar questions

H C I

p H = 10

- 0.118 V

H_{2}

p H = 3

^{o}

\frac{2.303 R T}{F}

0.059 V

p H = 3

25^{\circ} C

2.303 \frac{R T}{F}

0.059 V

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

Join BYJU'S Learning Program