At what pH oxidation potential of water is $- 0.81 V$ ? ( Standard oxidation potential of water = -1.23 V)

Open in App

Solution

The half reacttion for the oxidation of water is $2 H_{2} O \to O_{2} + 4 H^{+} + 4 e^{-} .$
The standard oxidation potential is $- 1.23 V$ .
The Nernst equation for the electrode potential is, $E_{c e l l} = E_{c e l l}^{\circ} - \frac{0.059}{n} l o g Q .$
Substitute values in the above expression :
$- 0.81 = - 1.23 - \frac{0.059}{4} l o g [H^{+}]^{4} .$
Hence, $[H^{+}] = 1 \times 10^{- 7} M .$
The corresponding pH is $- log [H^{+}]$
$= - log (1 \times 10^{- 7}) = 7.$

Suggest Corrections

Similar questions

- 1.23 V

= 0 V

298 K

+ 1.23 V

p H

- 0.81 V

M | M^{n +} (0.2 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m)

25^{\circ} C

C u^{2 +} (a q)

+ 0.34 V

M | M^{n +} (0.02 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m), P t

25^{o} C

0.81 V

0.76 V

log 0.02 = - 1.7

Join BYJU'S Learning Program