Q48 / 180 · COMEDK 2025 (25 May Shift) · Chemistry

For the reaction, $2 \mathrm{MnO}_4^{-}+b \mathrm{C}_2 \mathrm{O}_4^{2-}+c \mathrm{H}^{+} \rightarrow x \mathrm{Mn}^{2+}+y \mathrm{CO}_2+z \mathrm{H}_2 \mathrm{O}$ Find $b, c, x, y$ and $z$

  • A
    $b=4, \quad c=12, \quad x=2, \quad y=8, \quad z=6$
  • B
    $b=5, \quad c=12, \quad x=2, \quad y=10, \quad z=6$
  • Correct answer
    C
    $b=5, \quad c=16, \quad x=2, \quad y=10, \quad z=8$
  • D
    $b=4, \quad c=16, \quad x=2, \quad y=8, \quad z=8$

Solution

The given redox reaction is $2 \mathrm{MnO}_4^{-} + b \mathrm{C}_2 \mathrm{O}_4^{2-} + c \mathrm{H}^{+} \rightarrow x \mathrm{Mn}^{2+} + y \mathrm{CO}_2 + z \mathrm{H}_2 \mathrm{O}$.

The oxidation half-reaction is $\mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 2 \mathrm{CO}_2 + 2e^{-}$.

The reduction half-reaction is $\mathrm{MnO}_4^{-} + 8 \mathrm{H}^{+} + 5e^{-} \rightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H}_2 \mathrm{O}$.

To balance the electrons, multiply the oxidation half-reaction by 5 and the reduction half-reaction by 2:

$5 \mathrm{C}_2 \mathrm{O}_4^{2-} \rightarrow 10 \mathrm{CO}_2 + 10e^{-}$

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

Adding the two half-reactions gives:

$2 \mathrm{MnO}_4^{-} + 5 \mathrm{C}_2 \mathrm{O}_4^{2-} + 16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+} + 10 \mathrm{CO}_2 + 8 \mathrm{H}_2 \mathrm{O}$

Comparing this with the given equation, we find $b=5$, $c=16$, $x=2$, $y=10$, and $z=8$.

Answer: $b=5, \quad c=16, \quad x=2, \quad y=10, \quad z=8$

Hence, the correct answer is option: 'C'.

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