Question

Hydrogen sulphide (H2S) is a strong reducing agent.
Which of the following reactions shows its reducing
action -
(a) Cd(NO3)2 + H2S- CdS+2HNO3
(b) CuSO4 + HS - Cus + H2SO4
(c) 2FeCl3 +H2S - 2FeCl2+2HCI+S
(d) Pb(NO3)2 +H2S - PBS + 2CH3COOH
​

Answer 1

answer : option (c)

explanation : we know , reducing agent oxidizes itself and reduces other.

in option (c) ; $2FeCl_3+H_2S\Leftrightarrow 2FeCl_2+2HCl+S$

here, oxidation number or O.N of Fe in FeCl3 = +3

O.N of S in H2S = -2

O.N kf Fe in FeCl2 = +2

Oxidation number of S = 0

it is clear that, Fe is reduced from +3 to +2 whereas S is oxidised from -2 to O.

hence, this reaction shows that H2S is strong reducing agent.

hence, option (C) is correct choice.

Answer 2

Answer:

Explanation:

Option "C"

Hope you must hv got it....