fe2+ +mno4- - fe3+ +mn2+
fe2+ +mno4- - fe3+ +mn2+
You need to find the caps key and learn how to use it. I assume you want this balanced. I assume this is n acid solution from the permanganate half reaction.
5Fe^2+ + MnO4- + 8H+ ==> 5Fe^3+ +Mn^2+ + 4H2O
The chemical equation you provided represents a redox reaction between Fe2+ (iron in its +2 oxidation state) and MnO4- (permanganate ion) to form Fe3+ (iron in its +3 oxidation state) and Mn2+ (manganese in its +2 oxidation state).
To balance this equation, you need to ensure that the number of atoms and the total charge on both sides of the equation are the same. Here's how you can balance it step by step:
1. Write down the unbalanced equation:
Fe2+ + MnO4- → Fe3+ + Mn2+
2. Separate the equation into half-reactions, one for the oxidation half (loss of electrons) and one for the reduction half (gain of electrons):
Oxidation half-reaction: Fe2+ → Fe3+
Reduction half-reaction: MnO4- → Mn2+
3. Balance the atoms in each half-reaction, starting with elements other than oxygen and hydrogen. In the oxidation half-reaction, there is only one iron (Fe) atom on both sides, so no further balancing is needed for this element. In the reduction half-reaction, there is one manganese (Mn) atom on both sides too.
4. Balance the oxygen atoms by adding water (H2O) molecules to the side that needs more oxygen atoms. In this case, the reduction half-reaction needs balancing for oxygen. There are 4 oxygen atoms in MnO4-, so add 4 water molecules on the opposite side to balance the oxygen. Now the reduction half-reaction becomes:
MnO4- + 4H2O → Mn2+
5. Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs more hydrogen atoms. In this case, both half-reactions need balancing for hydrogen. The oxidation half-reaction has no hydrogen atoms, so add 8 hydrogen ions on the opposite side to balance the hydrogen. The reduction half-reaction already has enough hydrogen atoms from the water molecules.
Oxidation half-reaction: Fe2+ → Fe3+ + 8H+
Reduction half-reaction: MnO4- + 4H2O → Mn2+
6. Balance the charges in each half-reaction by adding electrons (e-) to the side that needs balancing. In the oxidation half-reaction, the charge increases from +2 to +3, so add one electron on the same side to balance the charge. In the reduction half-reaction, the charge decreases from +7 to +2 (from MnO4- to Mn2+), indicating that 5 electrons are gained. Add 5 electrons to the same side to balance the charge. Now both half-reactions are balanced in terms of atoms and charge.
Oxidation half-reaction: Fe2+ → Fe3+ + 8H+ + e-
Reduction half-reaction: MnO4- + 4H2O + 5e- → Mn2+
7. Multiply each half-reaction by coefficients so that both half-reactions have the same number of electrons. In this case, if we multiply the oxidation half-reaction by 5, we will have 5e- on both sides. The half-reactions become:
5Fe2+ → 5Fe3+ + 40H+ + 5e-
MnO4- + 4H2O + 5e- → Mn2+
8. Add the half-reactions together and cancel out the common terms (electron and hydrogen ions):
5Fe2+ + MnO4- + 4H2O → 5Fe3+ + Mn2+ + 8H+
Now the equation is balanced, and the coefficients represent the stoichiometric ratios of the reactants and products.