Acidified water is electrolysed with platinum electrodes at a steady current of $5.00 A$ for $1.00 h$ . The mass of water decomposed is (take $F = 96500 C mo l^{- 1}$ )

Answer: B

Charge

Q = 5.00 \times 3600 = 18000 C

, i.e.

18000/96500 = 0.1865 mol

of electrons. Overall

2 H X_{2} O 2 H X_{2} + O X_{2}

requires 4 electrons per 2 mol water, i.e. 2 electrons per mol water. Water decomposed

= 0.1865/2 = 0.0933 mol = 0.0933 \times 18 = 1.68 g