Manganate (VII) (MnO4-) ions react with Fe2+ ions in the presence of acid (H+) to form Mn2+ ions, Fe3+ ions and water. Write the overall redox equation for this reaction
(MnO4-)+(5Fe2+)+8H+ = (Mn2+)+(5Fe3+)+4H2O
That looks OK to me.
To write the overall redox equation for this reaction, we need to identify the oxidation and reduction half-reactions involved.
1. Oxidation half-reaction: In this reaction, Fe2+ ions are being oxidized to Fe3+ ions. The Fe2+ ions are losing electrons, so the oxidation half-reaction is:
Fe2+ → Fe3+ + e-
2. Reduction half-reaction: In this reaction, MnO4- ions are being reduced to Mn2+ ions. The MnO4- ions are gaining electrons, so the reduction half-reaction is:
MnO4- + 8H+ + 5e- → Mn2+ + 4H2O
Now, we need to balance the two half-reactions to ensure the number of electrons gained in the reduction half-reaction is equal to the number of electrons lost in the oxidation half-reaction.
Balancing the electrons:
The oxidation half-reaction (1) needs to be multiplied by 5 to balance the number of electrons:
5Fe2+ → 5Fe3+ + 5e-
The reduction half-reaction (2) already has 5 electrons in its equation.
Now, we can combine the two half-reactions to form the overall redox equation:
5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O
To write the overall redox equation for this reaction, we need to balance the oxidation and reduction half-reactions separately.
First, let's determine the oxidation half-reaction by observing the change in oxidation states for manganese (Mn). In the reactant, the oxidation state of Mn is +7, and in the product, it is reduced to +2. Therefore, the oxidation half-reaction is:
MnO4- → Mn2+
Next, let's determine the reduction half-reaction by observing the change in oxidation states for iron (Fe). In the reactant, the oxidation state of Fe is +2, and in the product, it is oxidized to +3. Therefore, the reduction half-reaction is:
Fe2+ → Fe3+
Since it is mentioned that the reaction occurs in the presence of acid, we can assume the presence of H+ ions as well. H+ ions are not involved in any change of oxidation states, so they can be added to both sides of the equation as needed to balance the charges.
Now let's combine the two half-reactions and balance them. Multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2 to ensure an equal number of electrons transferred in both half-reactions:
5(MnO4-) + 8H+ + 10e- → 5(Mn2+) + 4H2O (oxidation half-reaction)
2(Fe2+) → 2(Fe3+) + 2e- (reduction half-reaction)
To cancel out the electrons, we can multiply the reduction half-reaction by 5 and the oxidation half-reaction by 2:
10(Fe2+) → 10(Fe3+) + 10e-
Now we can add the resulting balanced half-reactions:
5(MnO4-) + 8H+ + 10(Fe2+) → 5(Mn2+) + 4H2O + 10(Fe3+)
Finally, simplify the equation by canceling out any common species on both sides:
2MnO4- + 16H+ + 10Fe2+ → 2Mn2+ + 8H2O + 5Fe3+
So, the overall redox equation for the reaction is:
2MnO4- + 16H+ + 10Fe2+ → 2Mn2+ + 8H2O + 5Fe3+