MnO4– + SO32– + H2O → MnO2 + SO42– + OH–
If the reaction occurs in an acidic solution where water and H+ ions are added to the half-reactions to balance the overall reaction, how many electrons are transferred in the balanced reduction half-reaction?
In an acidic solution the MnO4^- goes to Mn^2+ which is a gain of 5 electrons per mole MnO4^-. S in sulfite is 4+ and it changes to 6+ in sulfate which is a loss of 2 electrons/mol. After multiplying the Mn by 2 and the sulfite by 5 the total electons changes = 10.
So the reduction half (the MnO4^-) gains 5e/mol or 10e/2 mols in the balanced equation.
To determine the number of electrons transferred in the balanced reduction half-reaction, we need to identify the reduction half-reaction in the given chemical equation.
In this equation, MnO4– is being reduced to MnO2. The balanced reduction half-reaction can be written as follows:
MnO4– + 4H+ + 3e– → MnO2 + 2H2O
In this reduction half-reaction, we can see that 3 electrons (3e–) are being transferred.
Therefore, the number of electrons transferred in the balanced reduction half-reaction is 3.
To determine the number of electrons transferred in the balanced reduction half-reaction, we need to compare the oxidation states of the elements involved in the reaction before and after the transfer of electrons.
In MnO4–, the oxidation state of Mn is +7, and in MnO2, the oxidation state of Mn is +4. Therefore, there is a reduction of Mn from +7 to +4.
To balance the reduction half-reaction, we need to add electrons to the reactants to match the reduction in oxidation state.
In this case, to balance the reduction of Mn from +7 to +4, we need to add 3 electrons.
Therefore, 3 electrons are transferred in the balanced reduction half-reaction.