Electrochemistry

Question

135 gram of water is electrolysed during an experiment. if the efficiency of current is 75% then select the correct statement. (1) 168 L of O2 will be evolved at anode at STP (2) 84 L of H2 will be evolved at cathode at STP (3) 15 F electricity will be consumed (4) 20 F electricity will be consumed

Answer 1

Correct option is (4) 20 F electricity will be consumed

\(H_2 O \rightarrow H_2 + \frac {1}{2} O_2\)

i.e., 1 mol of \(H_2 O \rightarrow 1\) mol of H₂ and \(\frac {1}{2}\) mol of O₂

\(\therefore \frac {135}{18}\) = 7.5 mol of H₂ = 7.5 x 22.4 = 168 L

and \(\frac {7.5}{2}\) mol of O₂ = \(\frac {7.5}{2} \times 22.4 =\) 84 L

Now, 1 mol of H₂O ⟶ 2F of e^-

But efficiency = 75%

\(\therefore\) For 7.5 mol ⟶ 20F of e^-