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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.
Loss of electrons
Gain of electrons
MnO4- (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+ + 4H2O
3. Use H+ (as under acidic conditions) to then
balance the other side of the equation MnO4- + 8H+ ------> Mn2+ + 4H2O
4. Add electrons to balance the charges MnO4- + 8H+ + 5e- ------> Mn2+ + 4H2O
For the oxidation half equation,
1. Write out the reactant and product Fe2+ -------> 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+ + 4H2Osee more