In the following reaction what will be the ratio of coefficient of MnO4– and Fe+2 after balancing respectively MnO4– + Fe+2 → Mn+2 + Fe+31 : 21 : 31 : 41 : 5

To balance the given redox reaction, we first need to balance the atoms and charges in each half-reaction.

The half-reactions are:

MnO4– → Mn+2

Fe+2 → Fe+3

1. Balance the atoms other than O and H in each half-reaction:

For MnO4– → Mn+2, the Mn atoms are already balanced.

For Fe+2 → Fe+3, the Fe atoms are already balanced.

2. Balance the O atoms by adding H2O:

For MnO4– → Mn+2, add 4 H2O to the right side: MnO4– → Mn+2 + 4H2O

The Fe+2 → Fe+3 half-reaction does not contain O atoms.

3. Balance the H atoms by adding H+:

For MnO4– → Mn+2 + 4H2O, add 8 H+ to the left side: MnO4– + 8H+ → Mn+2 + 4H2O

The Fe+2 → Fe+3 half-reaction does not contain H atoms.

4. Balance the charges by adding electrons (e–):

For MnO4– + 8H+ → Mn+2 + 4H2O, add 5 e– to the right side: MnO4– + 8H+ + 5e– → Mn+2 + 4H2O

For Fe+2 → Fe+3, add 1 e– to the right side: Fe+2 → Fe+3 + e–

5. Make the electron gain in the two half-reactions equal:

Multiply the MnO4– + 8H+ + 5e– → Mn+2 + 4H2O half-reaction by 1 and the Fe+2 → Fe+3 + e– half-reaction by 5:

MnO4– + 8H+ + 5e– → Mn+2 + 4H2O

5Fe+2 → 5Fe+3 + 5e–

6. Add the half-reactions together and simplify:

MnO4– + 8H+ + 5Fe+2 → Mn+2 + 4H2O + 5Fe+3

So, the balanced equation is:

MnO4– + 8H+ + 5Fe+2 → Mn+2 + 4H2O + 5Fe+3

Therefore, the ratio of the coefficients of MnO4– and Fe+2 is 1:5.