JEE Advance - Chemistry (1983 - No. 1)

To complete and balance the given redox reaction between cerium(III) ions ($$Ce^{3+}$$) and peroxodisulfate ions ($$S_2O_8^{2-}$$) to form sulfate ions ($$SO_4^{2-}$$) and cerium(IV) ions ($$Ce^{4+}$$), we need to follow a step-by-step approach. This involves balancing the atoms and charges through oxidation and reduction half-reactions.

First, we separate the reaction into half-reactions:

Oxidation half-reaction (cerium is oxidized):

$$Ce^{3+} \to Ce^{4+} + e^-$$

Reduction half-reaction (peroxodisulfate is reduced):

$$S_2O_8^{2-} + 2e^- \to 2SO_4^{2-}$$

Next, we balance the electrons between the two half-reactions. The oxidation half-reaction releases 1 electron, while the reduction half-reaction consumes 2 electrons. To balance the electrons, we multiply the oxidation half-reaction by 2:

Oxidation half-reaction balanced with electrons:

$$2Ce^{3+} \to 2Ce^{4+} + 2e^-$$

Now, the electrons in both half-reactions are balanced, and we can combine them:

$$2Ce^{3+} + S_2O_8^{2-} \to 2Ce^{4+} + 2SO_4^{2-}$$

Finally, we have a balanced redox reaction. Both the number of atoms and the charges are balanced:

$$2Ce^{3+} + S_2O_8^{2-} \to 2Ce^{4+} + 2SO_4^{2-}$$

This reaction shows that two cerium(III) ions are oxidized to cerium(IV) ions, while a peroxodisulfate ion is reduced to two sulfate ions, with the transfer of two electrons to maintain charge balance.