Balance the following net iconic reactions, and identify which elements are oxidized and which are reduced:

a. MnO4^- (aq) + S^2- (aq) --> MnO2(s) + S(s)
b. IO3^- (aq) + I^- (aq) --> I2(s)
c. Mn^2+ (aq) + BiO3^- (aq) --> MnO4^- (aq) + Bi^3+ (aq)

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To balance these net iconic reactions and identify the elements that are oxidized and reduced, you'll need to follow these steps:

a. MnO4^- (aq) + S^2- (aq) --> MnO2(s) + S(s)

To balance the equation, first, balance the atoms other than hydrogen and oxygen. The sulfate ion (SO4^2-) has four oxygens, so we need to balance it by multiplying MnO4^- on the left by 4:

4 MnO4^- (aq) + S^2- (aq) --> MnO2(s) + S(s)

Next, balance the oxygens by adding water (H2O) on the right-hand side:

4 MnO4^- (aq) + S^2- (aq) --> MnO2(s) + S(s) + 4 H2O(l)

Now, balance the hydrogens by adding hydrogen ions (H+) on the left-hand side:

4 MnO4^- (aq) + S^2- (aq) + 8 H+ (aq) --> MnO2(s) + S(s) + 4 H2O(l)

Finally, balance the charge by adding electrons (e^-) on the left-hand side:

4 MnO4^- (aq) + S^2- (aq) + 8 H+ (aq) + 5 e^- --> MnO2(s) + S(s) + 4 H2O(l)

In this reaction, sulfur (S) gets oxidized (loses electrons) from -2 to 0, and manganese (Mn) gets reduced (gains electrons) from +7 to +4.

b. IO3^- (aq) + I^- (aq) --> I2(s)

To balance this equation, start by balancing the atoms other than hydrogen and oxygen. The iodine (I) atoms are already balanced, so we can move onto balancing the oxygens. Since IO3^- has three oxygens, we need to multiply I^- on the right by 3:

IO3^- (aq) + 3 I^- (aq) --> I2(s)

In this reaction, iodine (I) gets reduced (gains electrons) from -1 to 0.

c. Mn^2+ (aq) + BiO3^- (aq) --> MnO4^- (aq) + Bi^3+ (aq)

To balance this equation, first, balance the atoms other than hydrogen and oxygen. Since we have one manganese (Mn) and one bismuth (Bi) on each side, they are already balanced. Now, let's balance the oxygens by adding water (H2O):

Mn^2+ (aq) + BiO3^- (aq) --> MnO4^- (aq) + Bi^3+ (aq) + H2O(l)

Next, balance the hydrogens by adding hydrogen ions (H+):

Mn^2+ (aq) + BiO3^- (aq) + 4 H+ (aq) --> MnO4^- (aq) + Bi^3+ (aq) + H2O(l)

Finally, balance the charge by adding electrons (e^-):

Mn^2+ (aq) + BiO3^- (aq) + 4 H+ (aq) + 2 e^- --> MnO4^- (aq) + Bi^3+ (aq) + H2O(l)

In this reaction, bismuth (Bi) gets oxidized (loses electrons) from +5 to +3, and manganese (Mn) gets reduced (gains electrons) from +2 to +7.

Remember, oxidation involves the loss of electrons, and reduction involves the gain of electrons.