Chemistry, asked by ReddySuryaSuryareddy, 7 months ago

balance the following redox by ion-electron method
a)2
4 2 4 Mn aq HSO aq acidic medium 4M​

Answers

Answered by soumyaranjanjena22
1

Answer:

In the ion-electron method (also called the half-reaction method), the redox equation is separated into two half-equations - one for oxidation and one for reduction. Each of these half-reactions is balanced separately and then combined to give the balanced redox equation.

Answered by DollyRajoriya
1

Explanation:

The unbalanced chemical equation is:

MnO

4

(aq)+SO

2

(g)→Mn

2+

(aq)+HSO

4

(aq)

The oxidation half reaction is

SO

2

(g)+2H

2

O(l)→HSO

4

(aq)+3H

+

(aq)+2e

(aq).

The reduction half reaction is

MnO

4

(aq)→Mn

(

2+)(aq).

In the reduction half reaction, the oxidation number of Mn changes from +7 to +2. Hence, 5 electrons are added to LHS of the reaction.

MnO

4

(aq)+5e

→Mn

2+

(aq)

Charge is balanced in the reduction half reaction by adding 8 hydrogen ions to LHS.

MnO

4

(aq)+5e

+8H

+

(aq)→Mn

2+

(aq)

To balance O atoms, 4 water molecules are added on RHS.

MnO

4

(aq)+5e

+8H

+

(aq)→Mn

2+

(aq)+4H

2

O(l)

To equalize the number of electrons, the oxidation half reaction is multiplied by 5 and the reduction half reaction is multiplied by 2.

5SO

2

(g)+10H

2

O(l)→5HSO

4

(aq)+15H

+

(aq)+10e

(aq)

2MnO

4

(aq)+10e

+16H

+

(aq)→2Mn

2+

(aq)

Two half cell reactions are added to obtain a balanced equation.

2MnO

4

(aq)+5SO

2

(g)+2H

2

O(l)→2Mn

2+

(aq)+5HSO

4

(aq)

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