Q32 / 180 · COMEDK 2025 (10 May Shift 3) · Chemistry
Identify the correct coefficients (a), (b), (c) and (d) in the following equations
i) $\quad \mathrm{x} \mathrm{MnO}_4^{-}+$(a) $\mathrm{SO}_3{ }^{2-}+$ (b) $\mathrm{H}^{+} \longrightarrow \mathrm{x} \mathrm{Mn}^{2+}+$ (a) $\mathrm{SO}_4{ }^{2-}+$ (c) $\mathrm{H}_2 \mathrm{O}$
ii) (c) $\mathrm{S}_2 \mathrm{O}_3{ }^{2-}+$ (d) $\mathrm{MnO}_4{ }^{-}+\mathrm{H}_2 \mathrm{O} \rightarrow$ (d) $\mathrm{MnO}_2+$ (b) $\mathrm{SO}_4{ }^{2-}+\mathrm{x} \mathrm{OH}^{-}$
- A(a) =2 (b) =6 (c)=6 (d)=5
- B(a)=2 (b)=8 (c)=4 (d)=4
- Correct answerC(a)=5 (b)=6 (c)=3 (d)=8
- D(a)=3 (b)=8 (c)=6 (d)=6
Solution
For reaction (i): $x MnO_{4}^{-} + a SO_{3}^{2-} + b H^{+} \rightarrow x Mn^{2+} + a SO_{4}^{2-} + c H_{2}O$.
The oxidation half-reaction is $SO_{3}^{2-} + H_{2}O \rightarrow SO_{4}^{2-} + 2H^{+} + 2e^{-}$.
The reduction half-reaction is $MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$.
To balance electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. This gives $x=2$ and $a=5$.
The balanced equation is $2 MnO_{4}^{-} + 5 SO_{3}^{2-} + 6 H^{+} \rightarrow 2 Mn^{2+} + 5 SO_{4}^{2-} + 3 H_{2}O$.
Comparing coefficients, $a=5$, $b=6$, $c=3$.
For reaction (ii): $c S_{2}O_{3}^{2-} + d MnO_{4}^{-} + H_{2}O \rightarrow d MnO_{2} + b SO_{4}^{2-} + x OH^{-}$.
The oxidation half-reaction is $S_{2}O_{3}^{2-} + 10OH^{-} \rightarrow 2SO_{4}^{2-} + 5H_{2}O + 8e^{-}$.
The reduction half-reaction is $MnO_{4}^{-} + 2H_{2}O + 3e^{-} \rightarrow MnO_{2} + 4OH^{-}$.
Multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 8 gives $3 S_{2}O_{3}^{2-} + 8 MnO_{4}^{-} + H_{2}O \rightarrow 8 MnO_{2} + 6 SO_{4}^{2-} + 2 OH^{-}$.
Comparing coefficients with the given equation, $c=3$, $d=8$, $b=6$.
Thus, $a=5$, $b=6$, $c=3$, $d=8$.
Answer: (a)=5 (b)=6 (c)=3 (d)=8
The oxidation half-reaction is $SO_{3}^{2-} + H_{2}O \rightarrow SO_{4}^{2-} + 2H^{+} + 2e^{-}$.
The reduction half-reaction is $MnO_{4}^{-} + 8H^{+} + 5e^{-} \rightarrow Mn^{2+} + 4H_{2}O$.
To balance electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2. This gives $x=2$ and $a=5$.
The balanced equation is $2 MnO_{4}^{-} + 5 SO_{3}^{2-} + 6 H^{+} \rightarrow 2 Mn^{2+} + 5 SO_{4}^{2-} + 3 H_{2}O$.
Comparing coefficients, $a=5$, $b=6$, $c=3$.
For reaction (ii): $c S_{2}O_{3}^{2-} + d MnO_{4}^{-} + H_{2}O \rightarrow d MnO_{2} + b SO_{4}^{2-} + x OH^{-}$.
The oxidation half-reaction is $S_{2}O_{3}^{2-} + 10OH^{-} \rightarrow 2SO_{4}^{2-} + 5H_{2}O + 8e^{-}$.
The reduction half-reaction is $MnO_{4}^{-} + 2H_{2}O + 3e^{-} \rightarrow MnO_{2} + 4OH^{-}$.
Multiplying the oxidation half-reaction by 3 and the reduction half-reaction by 8 gives $3 S_{2}O_{3}^{2-} + 8 MnO_{4}^{-} + H_{2}O \rightarrow 8 MnO_{2} + 6 SO_{4}^{2-} + 2 OH^{-}$.
Comparing coefficients with the given equation, $c=3$, $d=8$, $b=6$.
Thus, $a=5$, $b=6$, $c=3$, $d=8$.
Answer: (a)=5 (b)=6 (c)=3 (d)=8
Hence, the correct answer is option: 'C'.