Consider the unbalanced redox reaction taking place in an acidic medium: \text{a MnO}_4^-(… — JAMB UTME

Consider the unbalanced redox reaction taking place in an acidic medium:

\text{a MnO}_4^-(\text{aq}) + \text{b SO}_3^{2-}(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d Mn}^{2+}(\text{aq}) + \text{e SO}_4^{2-}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest set of whole-number coefficients, what is the value of the stoichiometric coefficient

c

for

\text{H}^+(\text{aq})

Answer: 6

The value of the stoichiometric coefficient c for H+(aq) is 6.

Balancing the reduction half-reaction (

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

) and oxidation half-reaction (

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

) gives a combined total of

16\text{H}^+

on the reactant side and

10\text{H}^+

on the product side. Subtracting

10\text{H}^+

from both sides leaves a net coefficient of 6 for

\text{H}^+(\text{aq})

on the reactant side.

Write the balanced reduction half-reaction for permanganate ion in acidic medium.

\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

Manganese goes from oxidation state +7 to +2, requiring 5 electrons, 8 H+ ions to balance oxygen atoms, forming 4 H2O molecules.

Write the balanced oxidation half-reaction for sulfite ion to sulfate ion.

\text{SO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{SO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2e^-

Sulfur goes from oxidation state +4 to +6, releasing 2 electrons and 2 H+ ions while consuming 1 H2O molecule.

Equalize the number of transferred electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

The least common multiple of 5 and 2 transferred electrons is 10.

Combine the half-reactions and subtract common species (

10e^-

10\text{H}^+

, and

5\text{H}_2\text{O}

) from both sides.

2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{SO}_4^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

Subtracting

10\text{H}^+

from

16\text{H}^+

leaves

6\text{H}^+

on the reactant side, giving

c = 6

Balancing Redox Equations using the Ion-Electron Method in Acidic Medium

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