Calculate the $p H$ at which the potential of hydrogen electrode will be $0.059 V$ .

Open in App

Solution

we have,
$2 H^{+}$ + $2 e^{- 1}$ $\to$ $H_{2}$
$E_{c e l l}$ = $E_{c e l l}^{0}$ - 0.059 $l o g$ $\frac{1}{H^{+}}$

$E_{c e l l}$ = $E_{c e l l}^{0}$ - 0.059 $p H$
0.059 = 0.0 - 0.059 $p H$
$p H$ = -1

Suggest Corrections

Similar questions

p H = 3

^{o}

\frac{2.303 R T}{F}

0.059 V

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

p H = 3

25^{\circ} C

2.303 \frac{R T}{F}

0.059 V

N i / N i^{+ 2}

0.236 V

p H

25^{\circ} C

[N i^{+ 2}] = 1 M, \frac{R T}{F} = 0.059 V

Join BYJU'S Learning Program