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

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

Options

A. 67.2 litres
B. 68.1 litres
C. 73.2 litres
D. 74.1 litres

Answer

D

Explanation

Reduction of `H_2O` takes place at cathode since reduction potential of `H_2Ogt` reduction potential of `K^(o+)` ions.
At anode 2 mole of `CO_2` + 1 mol of propyne gas is obtained by 2 faraday (since two `bare` are produced).
2 faraday `implies` 3 mol of gases under standard conditions (1 bar, 298K)
`implies 3xx 24.7` litres `=74.1` litres
So the answer is (d).