Acidified water is electrolysed with platinum electrodes at a steady current of $2.00 A$ . The dry oxygen collected at the anode occupies $492 mL$ at $300 K$ and $0.960 atm$ . Find the time of electrolysis in seconds (nearest integer). Take $R = 0.0821 L atm K^{- 1} mo l^{- 1}$ , $F = 96500 C mo l^{- 1}$ .

Numerical answer type, answer in s, tolerance 3.

Answer: 3701 s

First get moles of

O X_{2}

from the gas law:

n = \frac{P V}{RT} = \frac{0.960 \times 0.492}{0.0821 \times 300} = 0.01918 mol

. Anode:

2 H X_{2} O O X_{2} + 4 H X^{+} + 4 e X^{-}

, so electrons

= 4 \times 0.01918 = 0.07671 mol

and

Q = 0.07671 \times 96500 = 7402 C

. Time

t = Q / I = 7402/2.00 = 3701 s