In hot, concentrated alkaline solutions, chlorine gas undergoes a disproportionation redox… — JAMB UTME

In hot, concentrated alkaline solutions, chlorine gas undergoes a disproportionation redox reaction according to the unbalanced equation:

\text{Cl}_2(\text{g}) + \text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + \text{Cl}^-(\text{aq}) + \text{H}_2\text{O}(\text{l})

When this equation is balanced using the smallest set of whole-number coefficients, what is the stoichiometric coefficient of hydroxide ions (

\text{OH}^-

) and the total number of moles of electrons transferred in the balanced equation?

The balanced equation requires 6 hydroxide ions and involves the transfer of 5 moles of electrons.

In the balanced redox equation

3\text{Cl}_2(\text{g}) + 6\text{OH}^-(\text{aq}) \rightarrow \text{ClO}_3^-(\text{aq}) + 5\text{Cl}^-(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

, the stoichiometric coefficient of hydroxide ions is 6, and 5 moles of electrons are transferred per mole of reaction as written.

Assign oxidation numbers to determine the oxidation and reduction half-reactions.

Elemental chlorine

\text{Cl}_2

has an oxidation number of

0

. In

\text{ClO}_3^-

, chlorine has an oxidation state of

+5

(oxidation). In

\text{Cl}^-

, chlorine has an oxidation state of

-1

(reduction).

Disproportionation involves the simultaneous oxidation and reduction of the same element.

Write and balance the oxidation half-reaction in basic medium.

\frac{1}{2}\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 3\text{H}_2\text{O} + 5e^-

One chlorine atom increases in oxidation state from

0

+5

, releasing

5e^-

. Six

\text{OH}^-

ions balance the charge and oxygen/hydrogen mass.

Write and balance the reduction half-reaction.

\frac{1}{2}\text{Cl}_2 + e^- \rightarrow \text{Cl}^-

One chlorine atom decreases in oxidation state from

0

-1

, accepting

1e^-

Equalize electron transfer between half-reactions and combine.

Multiply the reduction half-reaction by 5:

\frac{5}{2}\text{Cl}_2 + 5e^- \rightarrow 5\text{Cl}^-

. Combine with the oxidation half-reaction:

3\text{Cl}_2 + 6\text{OH}^- \rightarrow \text{ClO}_3^- + 5\text{Cl}^- + 3\text{H}_2\text{O}

. Total electrons transferred

n_e = 5

The number of electrons lost in oxidation must equal the number gained in reduction.

Balancing Disproportionation Redox Reactions in Basic Medium

Question