Balance: Mno4 + Fe2+ - Fe3+ +Mn2+
[MnO4]^- + 5Fe^2+ + 8H^+ ==> 5Fe^3+ + Mn^2+ + 4H2O
The given balanced chemical equation is:
MnO4- + Fe2+ → Fe3+ + Mn2+
To balance this chemical equation, we need to ensure that the number of each type of atom is the same on both sides of the equation.
Let's start by balancing the atoms that appear in multiple species.
From the left-hand side, we have one manganese (Mn) atom and four oxygen (O) atoms.
From the right-hand side, we have one manganese (Mn) atom and four oxygen (O) atoms.
Since there is an equal number of these atoms on both sides already, we don't need to adjust them further.
Next, let's balance the electrons (e-) in the equation:
The MnO4- ion has a charge of -1, and the Fe2+ ion has a charge of +2. Since the overall charge must be balanced, we need to multiply the MnO4- ion by two to get two electrons on the left-hand side:
2 MnO4- + Fe2+ → Fe3+ + Mn2+ + 2e-
Now, let's balance the charges:
The Fe2+ ion has a charge of +2, but on the right-hand side, it is converted to Fe3+, which has a charge of +3. To balance this, we can add one electron (e-) to the right-hand side:
2 MnO4- + Fe2+ → Fe3+ + Mn2+ + 2e- + e-
Finally, combining the electrons and simplifying:
2 MnO4- + Fe2+ → Fe3+ + Mn2+ + 3e-
This is the balanced chemical equation for the given reaction, where the numbers in front of each compound or species represent the stoichiometric coefficients.
The given chemical equation is:
MnO4^- + Fe^2+ -> Fe^3+ + Mn^2+
Step 1: Determine the oxidation numbers of the elements involved.
In MnO4^-, the oxidation number for Mn is +7 since oxygen has an oxidation number of -2.
In Fe^2+, the oxidation number for Fe is +2.
Step 2: Determine which elements are being oxidized and reduced.
In this reaction, Mn is being reduced because it is going from an oxidation number of +7 to +2.
Fe is being oxidized because it is going from an oxidation number of +2 to +3.
Step 3: Write the half-reactions for each process.
Reduction half-reaction: MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O
Oxidation half-reaction: Fe^2+ -> Fe^3+ + e^-
Step 4: Balance the number of electrons transferred in each half-reaction.
Multiply the oxidation half-reaction by 5 to balance the number of electrons:
5Fe^2+ -> 5Fe^3+ + 5e^-
Step 5: Combine the two half-reactions to form the balanced overall equation.
5Fe^2+ + MnO4^- + 8H+ -> 5Fe^3+ + Mn^2+ + 4H2O