In the following sequence of reactions, products (A) to (H) are formed:

The gases (B), (C), and (D), respectively, are:

Options

A. 11.2 litres
B. 11.35 litres
C. 22.4 litres
D. 22.7 litres

Answer

B

Explanation

Reduction of `H_2O` takes place at cathode since reduction potential of `H_2Ogt` reduction potential of `K^(o+)` ions.
At cathode 1 mol `H_2(g)` is obtained by 2 faraday (since two `bare`'s are used in the reduction).
2 faraday `implies` 1 mol of `H_2` at STP ( 1 bar and 273 K)
1 faraday `implies1/2` mol of `H_2implies1/2xx22.7=11.35` litres
So the answer is (b).