Question

Balance the following reaction by Oxidation number
method.

pls answer the question pls...​

Answer 1

Answer

(a) Write unbalance equation 2S−2+MnO4−→S0+M+2n2++H2O

(b) Oxidation number of Mn decreases from +7 to +2. So, MnO4− is reduced to Mn+2 and oxidation number of S increase from −2 to 0 so, H2S is oxidised to S. Hence writing partial equation as, 

M+7nO4−→M2+n2+ 

Oxidation number decreases 5 (Reduction)

H2S−2→S0

Oxidation number increases by 2

Multiplying equation 2 and equation 3 by 5, we get

2MnO4−→2Mn+2

5H2S→5S

Adding equation 4 and 5 we get,

2MnO4−+5H2S→2Mn2++5S

Change (−1)(0)(+4)(0)

(c) Adding of water molecules in the direction of deficiency of oxygen

2MnO4+5H2S→2