Write half-reactions for the oxidation and reduction processes for the following:
Sn2+ + IO3- -> Sn4+ + I- (acidic solution)
MN2+ + H2O2 -> MnO2 + H2O (basic solution)
To write half-reactions for oxidation and reduction processes, we need to identify the changes in oxidation states of the elements involved and balance the atoms and charges in the reaction.
Let's start with the first equation in an acidic solution:
Sn2+ + IO3- -> Sn4+ + I-
1. Oxidation: In this reaction, the Sn2+ ion is being oxidized to Sn4+. The oxidation state of tin (Sn) increases from +2 to +4.
Sn2+ -> Sn4+ (Oxidation half-reaction)
2. Reduction: In this reaction, the IO3- ion is being reduced to I-. The oxidation state of iodine (I) decreases from +5 to -1.
IO3- -> I- (Reduction half-reaction)
Now, let's consider the second equation in a basic solution:
Mn2+ + H2O2 -> MnO2 + H2O
1. Oxidation: In this reaction, the Mn2+ ion is being oxidized to MnO2. The oxidation state of manganese (Mn) increases from +2 to +4.
Mn2+ -> MnO2 (Oxidation half-reaction)
2. Reduction: In this reaction, the H2O2 molecule is being reduced to H2O. The oxidation state of oxygen (O) decreases from -1 to -2.
H2O2 -> H2O (Reduction half-reaction)
Remember, these half-reactions are not balanced yet. You can balance them by adding appropriate coefficients to ensure that the number of atoms and charge are equal on both sides of each half-reaction.