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

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

Options

A. `{:(ul(" (B) (C) (D)")),((a)""HC-=CH" "CO_(2)" "H_(2)):}`
B. `{:(ul(" (B) (C) (D)")),((a)""HC-=CH" "H_(2)" "CO_(2)):}`
C. `{:(ul(" (B) (C) (D)")),((a)""Me--=-Me" "CO_(2)" "H_(2)):}`
D. `{:(ul(" (B) (C) (D)")),((a)""Me--=-H" "H_(2)" "CO_(2)):}`

Answer

C

Explanation

Reduction of `H_2O` takes place at cathode since reduction potential of `H_2Ogt` reduction potential of `K^(o+)` ions.
Gases B, C, D (propyne, `CO_2` and `H_2`, respectively), answer is (c).