Mno4 + fe2 ----> mn2 + fe3
Balance the redox reaction in acidic medium
MnO4^- + 5Fe^2+ + 8H^+ ---> Mn^2+ + 5Fe^3+ + 4H2O
To balance the given redox reaction, we can follow these steps:
Step 1: Write the skeleton equation:
MnO4^- + Fe^2+ -> Mn^2+ + Fe^3+
Step 2: Identify the oxidation numbers of each element:
In this reaction, the oxidation number of Mn changes from +7 to +2, and the oxidation number of Fe changes from +2 to +3.
Step 3: Divide the reaction into two half-reactions, one for the oxidation and one for the reduction:
Oxidation half-reaction: MnO4^- -> Mn^2+
Reduction half-reaction: Fe^2+ -> Fe^3+
Step 4: Balance the atoms other than hydrogen and oxygen in each half-reaction:
Oxidation half-reaction: 4 H+ + MnO4^- -> Mn^2+ (Note that there are now 4 hydrogen ions on the left side)
Reduction half-reaction: Fe^2+ -> Fe^3+
Step 5: Balance the oxygen atoms by adding water molecules to the side that needs them:
Oxidation half-reaction: 4 H+ + MnO4^- -> Mn^2+ + 4 H2O (Note that there are now 4 oxygen atoms on the right side)
Reduction half-reaction: Fe^2+ -> Fe^3+
Step 6: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that needs them:
Oxidation half-reaction: 4 H+ + MnO4^- -> Mn^2+ + 4 H2O
Reduction half-reaction: 2 Fe^2+ -> 2 Fe^3+ + 2 H+
Step 7: Balance the charges by adding electrons (e-) to the side that needs them:
Oxidation half-reaction: 4 H+ + MnO4^- + 3e- -> Mn^2+ + 4 H2O
Reduction half-reaction: 2 Fe^2+ -> 2 Fe^3+ + 2 H+ + 2e-
Step 8: Multiply each half-reaction by a number so that the number of electrons gained in the reduction equals the number of electrons lost in the oxidation. In this case, we need to multiply the oxidation half-reaction by 2 and the reduction half-reaction by 3:
2(4 H+ + MnO4^- + 3e-) -> 2(Mn^2+ + 4 H2O)
3(2 Fe^2+ -> 2 Fe^3+ + 2 H+ + 2e-)
Step 9: Add the two half-reactions together and cancel out common terms:
8 H+ + 2 MnO4^- + 6e- + 6 Fe^2+ -> 2 Mn^2+ + 4 H2O + 6 Fe^3+
Finally, the balanced redox reaction in acidic medium is:
8 H+ + 2 MnO4^- + 6 Fe^2+ -> 2 Mn^2+ + 4 H2O + 6 Fe^3+
To balance the redox reaction in acidic medium, we can follow these steps:
Step 1: Write the unbalanced equation:
MnO4^- + Fe^2+ -> Mn^2+ + Fe^3+
Step 2: Separate the half-reactions for oxidation and reduction.
Oxidation half-reaction:
Fe^2+ -> Fe^3+
Reduction half-reaction:
MnO4^- -> Mn^2+
Step 3: Balance the atoms other than hydrogen and oxygen in each half-reaction.
In the oxidation half-reaction, there is only one iron (Fe) atom on both sides, so it is already balanced.
In the reduction half-reaction:
MnO4^- -> Mn^2+
There is one manganese (Mn) atom on both sides.
But there are four oxygen (O) atoms on the left side (MnO4^-) and only two oxygen (O) atoms on the right side (Mn^2+). So, we need to balance the oxygen atoms.
We can add water (H2O) to the side that has the fewer oxygen atoms. In this case, we add four water molecules to the right side:
MnO4^- + 4H2O -> Mn^2+
Now, there are four manganese (Mn) atoms on both sides.
Step 4: Balance the hydrogen atoms by adding hydrogen ions (H+) to the side that lacks hydrogen.
MnO4^- + 4H2O -> Mn^2+ + 8H+
Now, there are eight hydrogen (H) atoms on the right side.
Step 5: Balance the charges by adding electrons (e^-) to the side that has a higher positive charge.
MnO4^- + 4H2O + 5e^- -> Mn^2+ + 8H+
In the oxidation half-reaction, we need to balance the electrons.
Fe^2+ -> Fe^3+ + e^-
Step 6: Multiply the half-reactions by appropriate coefficients so that the number of electrons generated in the oxidation half-reaction is equal to the number of electrons consumed in the reduction half-reaction.
In this case, we need 5 electrons in the oxidation half-reaction and 5 electrons in the reduction half-reaction to balance the electron transfers.
5(Fe^2+ -> Fe^3+ + e^-)
MnO4^- + 4H2O + 5e^- -> Mn^2+ + 8H+
Step 7: Add the balanced half-reactions together.
5Fe^2+ + MnO4^- + 4H2O -> 5Fe^3+ + Mn^2+ + 8H+
Finally, the balanced redox reaction in acidic medium is:
5Fe^2+ + MnO4^- + 4H2O -> 5Fe^3+ + Mn^2++ 8H+