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

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

Options

A. 1
B. 13
C. 2
D. 12

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, 2 mol of `overset(Ө)OH` is obtained by 2 faraday.
(Or 1 faraday produces 1 equivalent of `overset(Ө)OH` ions.)
faraday `implies` 1 mol of `overset(Ө)OH` ions
Concentration of `overset(Ө)OH` ions =Mole/Volume`=(1)/(10)`
`=0.1M=10^-1M`
`pOH=-log[OH]=-log[10^(-1)]=1`
`pH=14-pOH=14-1=13`
So the answer is (b).