How to formulate and balance a redox equation under acidic conditions

In a reaction between  MnO₄^- and Fe²⁺ we are told MnO₄^- goes to Mn²⁺ and Fe²⁺ goes to Fe³⁺. We then need to derive the redox equation for the reaction occuring. First we need to idenfity which species are being reduced and which are oxidised. We can use ' OIL RIG' when referring to electrons to help us.OxidationIsLoss of electronsReduction Is Gain of electronsMnO₄^- (where Mn has a +7 oxidation state) gains electrons, hence is reduced, to Mn²⁺ and Fe²⁺ loses electrons, hence oxidised, to Fe³⁺. Starting with the reduction half equation,    1. Write out the reactant and product                   MnO₄^{-        ------>}   Mn²⁺2. Use H₂O to balance the oxygen                    MnO₄^{-        ------>}   Mn²⁺ + 4H₂O3. Use H⁺ (as under acidic conditions) to thenbalance the other side of the equation                  MnO₄- + 8H⁺        ------>   Mn²⁺ + 4H₂​O4. Add electrons to balance the charges                MnO₄- + 8H⁺ + 5e^-    ------>   Mn²⁺ + 4H₂​OFor the oxidation half equation,1. Write out the reactant and product                  Fe^{2+   ------->}   Fe³⁺2. Balance the equation using electrons                Fe²⁺ ------->   Fe³⁺ + e^-To combine the reduction and oxidation half equations we need to have the same number of electrons. Multiply the oxidation half equation by 5 to give us 5Fe²⁺  ------->   5Fe³⁺ + 5e^-Then add the two half equations together, canceling any like terms (the electrons) from both sides. The overall redox equation is therefore 5Fe²⁺ + MnO₄- + 8H⁺   ------>  5Fe³⁺ + Mn²⁺ + 4H₂​O