I m having a tough time balancing redox equations!?

Method 1 : Balancing the charges

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

When electrons have not yet been added to the equation :
Total charges on the left = (-1) + 8 × (+1) = +7
Total charges on the right = +2

To make the charges are equal on the both side, 5e⁻ should be added to the left, i.e.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O


Method 2 : By the change in oxidation number

MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Oxidation number of Mn on the left (in MnO₄⁻) = +7
Oxidation number of Mn on the right (in Mn²⁺) = +2

Decrease in oxidation number = 5
To decrease the oxidation number by 5, 5e⁻ should be added to the reactant side, i.e.
MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Add up the charges on each side using the numbers of each charge and multiplying by the numbers of ions.
Remember plusses cancel out minuses.
Then, write out the remaining numbers for each side using an = sign.
Now, add electrons to one side or other to balance the equation. .
MnO4^- + 8H^+ --> Mn^2+ + 4H20

1 - + 8+ = 2+
>>> 7+ = 2+ So we need to add 5e- to the left side

trying googling about equations