One purpose of adding KIO4 is to prevent the permanganate ion from being slowly reduced to Mn+2 by H2O. Write a balanced redox equation for the reduction of permanganate ion to Mn+2 by water in acid solution.
MnO4^- + 8H^+ + 5e ==> Mn^2+ + 4H2O
Don't memorize these half reactions; there are thousands. Instead, learn how to write them. Here is a site OR ask specific questions about what you may not understand.
http://www.chemteam.info/Redox/Redox.html
To write a balanced redox equation for the reduction of permanganate ion (MnO4-) to Mn+2 by water (H2O) in an acidic solution, you need to follow a few steps:
1. Start by writing down the unbalanced equation for the reaction:
MnO4- + H2O → Mn+2 + ?
2. Determine the oxidation states of each element and ion involved in the reaction:
The oxidation state of Mn in MnO4- is +7, since the overall charge of the ion is -1.
The oxidation state of H in H2O is +1, since hydrogen usually has an oxidation state of +1 in compounds.
The oxidation state of Mn in Mn+2 is +2, since the ion itself has a charge of +2.
3. Determine the changes in oxidation states for the atoms undergoing redox reactions:
The Mn atom is reduced from +7 to +2, so it gains 5 electrons.
The H atoms in water do not undergo any changes in their oxidation states, so they are not involved in the redox reaction.
4. Balance the number of electrons lost and gained:
Since one permanganate ion (MnO4-) is reduced to one Mn+2 ion, you need 5 electrons to balance the reduction.
5. Write the final balanced redox equation:
MnO4- + 8H+ + 5e- → Mn+2 + 4H2O
The final balanced redox equation shows the reduction of permanganate ion to Mn+2 by water in an acidic solution.