Balance the following redox reaction.
MnO4 (aq) + Fe (s) --> Mn2+ (aq) +Fe2+ (aq)
I expect this is just an exercise in balancing redox equations; from a practical standpoint I doubt there is much reaction between MnO4^- and solid Fe.
very bad
Why did the redox reaction cross the road? To find its equilibrium!
To balance this redox reaction, we need to make sure that the number of atoms and the charge are equal on both sides of the equation.
Let's start by balancing the Fe atoms:
Fe (s) --> Fe2+ (aq)
Now let's balance the Mn atoms:
MnO4 (aq) + Fe (s) --> Mn2+ (aq) + Fe2+ (aq)
Next, let's balance the oxygen atoms by adding water (H2O) molecules:
MnO4 (aq) + Fe (s) --> Mn2+ (aq) + Fe2+ (aq) + H2O
Now let's balance the hydrogen atoms by adding hydrogen ions (H+):
MnO4 (aq) + Fe (s) + 8H+ --> Mn2+ (aq) + Fe2+ (aq) + H2O
Finally, let's balance the charges by adding electrons:
MnO4 (aq) + 5Fe (s) + 8H+ + 5e- --> Mn2+ (aq) + 5Fe2+ (aq) + H2O
And there you have it! The balanced redox reaction:
MnO4 (aq) + 5Fe (s) + 8H+ --> Mn2+ (aq) + 5Fe2+ (aq) + H2O
To balance a redox reaction, follow these steps:
1. Determine the oxidation numbers for each element in the reaction.
MnO4: Mn has an oxidation number of +7 and O has an oxidation number of -2.
Fe: Fe has an oxidation number of 0.
Mn2+: Mn has an oxidation number of +2.
Fe2+: Fe has an oxidation number of +2.
2. Identify the elements that are changing oxidation numbers. In this case, Mn is being reduced from +7 to +2, and Fe is being oxidized from 0 to +2.
3. Write separate half-reactions for the oxidation and reduction processes.
Oxidation (loss of electrons):
Fe --> Fe2+ (oxidation number changes from 0 to +2)
Reduction (gain of electrons):
MnO4- --> Mn2+ (oxidation number changes from +7 to +2)
4. Balance the non-oxygen and non-hydrogen atoms in each half-reaction using coefficients.
Oxidation: Fe --> Fe2+ (already balanced)
Reduction: MnO4- --> Mn2+
There are four oxygen atoms on the reactant side, so add four water (H2O) molecules to the product side to balance oxygen.
MnO4- --> Mn2+ + 4H2O
5. Balance the hydrogen atoms by adding hydrogen ions (H+).
Oxidation: Fe --> Fe2+ (already balanced)
Reduction: MnO4- + 8H+ --> Mn2+ + 4H2O
6. Balance the charge in each half-reaction by adding electrons (e-). The total charge on each side should be equal.
Oxidation: Fe --> Fe2+ + 2e-
The oxidation half-reaction has a charge of 0 on both sides since the number of electrons is equal.
Reduction: MnO4- + 8H+ + 5e- --> Mn2+ + 4H2O
The reduction half-reaction has a charge of -1 on the reactant side (-1 from MnO4- and +8 from 8H+), and a charge of +2 on the product side (from Mn2+). To balance the charges, add 5 electrons to the reactant side.
7. Multiply the two half-reactions by coefficients so that the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction. In this case, multiply the oxidation half-reaction by 5, and the reduction half-reaction by 2.
Oxidation: 5Fe --> 5Fe2+ + 10e-
Reduction: 2MnO4- + 16H+ + 10e- --> 2Mn2+ + 8H2O
8. Add the two balanced half-reactions together.
5Fe + 2MnO4- + 16H+ --> 5Fe2+ + 2Mn2+ + 8H2O
first step: break down the equation into half equations for oxidation and reduction
reduction reaction;
8H+ + 5e- + MnO4- --> Mn2+ + 4H2O
oxidation reaction;
Fe --> Fe2+ + 2e-
second step: manipulate the equations in order to get the same number of electrons; i.e. multiply reduction equation with 2 to get;
16H+ + 10e- + 2MnO4- --> 2Mn2+ + 8H2O
multiply oxidation reaction with 5 to get;
5Fe --> 5Fe2+ + 10e-
third step: add the two equations together (cancel variables on different side of the equation)i.e. the electrons are cancelled out.
16H+ + 5Fe + 2MnO4- --> 2Mn2+ + 5Fe2+ + 8H2O