Chapter 13, Problem 3CO

• write and balance half-reactions for simple redox processes.

Interpretation Introduction

Interpretation:

Half reactions for a simple redox process should be written and balanced.

Concept introduction:

• Redox reaction is any chemical reaction in which one of the reactants is oxidized and the other one is reduced simultaneously.
• It involves the transfer of electrons between species.
• A redox reaction can be separated into two half-reactions: one representing oxidation of reactant and other representing reduction reaction of another reactant.

Answer to Problem 3CO

Solution: In case of $Mn{O}_4^{-}(aq)+I^{-}(aq)\to M{n}^{2+}(aq)+I_2(s)$

• Balanced half-reactions are:
• Oxidation half-reaction: $2I^{-}(aq)\to I_2(s)+2e^{-}$
• Reduction half-reaction: $Mn{O}_4^{-}(aq)+8H^{+}+5e^{-}\to M{n}^{2+}(aq)+4H_{2}O$

• And balanced redox reaction is:

  $2Mn{O}_4^{-}(aq)+16H^{+}+10I^{-}\to 2M{n}^{2+}(aq)+8H_{2}O+5I_2(s)$

Explanation of Solution

To balance a simple redox reaction, first separate it into two half-reactions, one for oxidation and other for reduction. Then each half-reaction is balanced following these basic rules:

1. Each element in the half-reaction is balanced except O and H.
2. Oxygen atoms are then balanced by adding the required number of water molecules (H₂O) to the opposite side of the equation.
3. Following O, hydrogen atom is balanced by adding H+ ions to the opposite side of the reaction
4. Due to the addition of H+ ions, the charge on both sides of the equation is not equal which are then balanced by adding enough electrons to the more positive side.
5. Finally, the half reactions are then added and the common terms are canceled out including electrons.

Example1: balance the following redox reaction:

$Mn{O}_4^{-}(aq)+I^{-}(aq)\to M{n}^{2+}(aq)+I_2(s)$

Step 1: Divide the redox reaction into two half reactions-

Oxidation state of Mn in $Mn{O}_4^{2-}$ is +7 and during the reaction it changes from +7 to +2. Hence, Mn gains electrons and is reduced. On the other hand, oxidation state of iodine changes from -1 to 0. Thus, it is loses electrons and is oxidised.

$\therefore$

Oxidation half-reaction: $I^{-}(aq)\to I_2(s)$

Reduction half-reaction: $Mn{O}_4^{-}(aq)\to M{n}^{2+}(aq)$

Step 2: Balance elements except O and H-

Oxidation half-reaction: $2I^{-}(aq)\to I_2(s)$

Reduction half-reaction: $Mn{O}_4^{-}(aq)\to M{n}^{2+}(aq)$

Step 3: Balance oxygen atoms by adding water molecules-

Oxidation half-reaction: $2I^{-}(aq)\to I_2(s)$

Reduction half-reaction: $Mn{O}_4^{-}(aq)\to M{n}^{2+}(aq)+4H_{2}O$

Step 4: Balance Hydrogen atom by adding H+ ions-

Oxidation half-reaction: $2I^{-}(aq)\to I_2(s)$

Reduction half-reaction: $Mn{O}_4^{-}(aq)+8H^{+}\to M{n}^{2+}(aq)+4H_{2}O$

Step 5: Balance the charge by adding electrons-

Oxidation half-reaction: $2I^{-}(aq)\to I_2(s)+2e^{-}$

Reduction half-reaction: $Mn{O}_4^{-}(aq)+8H^{+}+5e^{-}\to M{n}^{2+}(aq)+4H_{2}O$

Step 6: multiply the oxidation half by 5 and reduction half by 2 to make the number of electrons in both reactions equal-

Oxidation half-reaction: $5\times (2I^{-}(aq)\to I_2(s)+2e^{-})=10I^{-}(aq)\to 5I_2(s)+10e^{-}$

Reduction half-reaction:

$\begin{array}{l}2\times (Mn{O}_4^{-}(aq)+8H^{+}+5e^{-}\to M{n}^{2+}(aq)+4H_{2}O)\\ =2Mn{O}_4^{-}(aq)+16H^{+}+10e^{-}\to 2M{n}^{2+}(aq)+8H_{2}O\end{array}$

Step 7: Add both half reaction and cancel out the common terms to get the balanced redox reaction-

$\begin{array}{l}2Mn{O}_4^{-}(aq)+16H^{+}+10e^{-}\to 2M{n}^{2+}(aq)+8H_{2}O\\ 10I^{-}(aq)\to 5I_2(s)+10e^{-}\end{array}$

$2Mn{O}_4^{-}(aq)+16H^{+}+10I^{-}\to 2M{n}^{2+}(aq)+8H_{2}O+5I_2(s)$

Conclusion

The balanced redox reaction for the example is:

$2Mn{O}_4^{-}(aq)+16H^{+}+10I^{-}\to 2M{n}^{2+}(aq)+8H_{2}O+5I_2(s)$