Question:
Consider the following reaction:
\[
xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+} + qIO_3^- + rH_2O
\]
The correct ratio \(x:y\) in the balanced equation is:
Consider the following reaction:
\[
xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+} + qIO_3^- + rH_2O
\]
The correct ratio \(x:y\) in the balanced equation is:
Show Hint
In acidic medium redox reactions, first calculate the change in oxidation numbers. The ratio of reactants is often obtained quickly by equating total electrons lost and gained before completing the full balancing process.
Updated On: Jun 11, 2026
- \(6:5\)
- \(6:4\)
- \(1:1\)
- \(5:4\)
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The Correct Option is A
Solution and Explanation
Concept:
This is a redox reaction occurring in acidic medium. To balance such reactions accurately, we use the
ion-electron method (also known as the oxidation number method). The key idea is that the total number of electrons lost during oxidation must be equal to the total number of electrons gained during reduction. We first identify the oxidation and reduction half-reactions and then combine them appropriately.
Step 1: Identify the species undergoing oxidation and reduction. The reactants are: \[ MnO_4^-, \quad I^-, \quad H^+ \] The products are: \[ Mn^{2+}, \quad IO_3^-, \quad H_2O \] For manganese: \[ MnO_4^- \] Oxidation state of Mn: \[ x+4(-2)=-1 \] \[ x-8=-1 \] \[ x=+7 \] In the product: \[ Mn^{2+} \] Oxidation state = \(+2\). Thus manganese undergoes reduction: \[ Mn^{+7}\rightarrow Mn^{+2} \] Gain in electrons: \[ 5e^- \] Therefore, each permanganate ion gains \(5\) electrons.
Step 2: Determine oxidation of iodide ion. For iodide: \[ I^- \rightarrow IO_3^- \] Oxidation state of iodine in \(I^-\): \[ -1 \] Oxidation state of iodine in \(IO_3^-\): \[ x+3(-2)=-1 \] \[ x-6=-1 \] \[ x=+5 \] Thus iodine changes from \[ -1 \rightarrow +5 \] Increase in oxidation number: \[ 6 \] Hence each iodide ion loses \[ 6e^- \] during oxidation.
Step 3: Write the half-reactions. Reduction half-reaction: \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+}+4H_2O \] Oxidation half-reaction: \[ I^-+3H_2O \rightarrow IO_3^-+6H^++6e^- \]
Step 4: Equalize the electrons exchanged. The reduction half-reaction involves \[ 5e^- \] while the oxidation half-reaction involves \[ 6e^-. \] The least common multiple of \(5\) and \(6\) is \[ 30. \] Therefore, Multiply the reduction half-reaction by \(6\): \[ 6MnO_4^- +48H^+ +30e^- \rightarrow 6Mn^{2+}+24H_2O \] Multiply the oxidation half-reaction by \(5\): \[ 5I^-+15H_2O \rightarrow 5IO_3^-+30H^++30e^- \]
Step 5: Add the two balanced half-reactions. Adding and cancelling electrons: \[ 6MnO_4^-+5I^-+18H^+ \rightarrow 6Mn^{2+}+5IO_3^-+9H_2O \] Thus the balanced equation is: \[ 6MnO_4^-+5I^-+18H^+ \rightarrow 6Mn^{2+}+5IO_3^-+9H_2O \]
Step 6: Determine the required ratio. Comparing with \[ xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+}+qIO_3^-+rH_2O \] we obtain: \[ x=6 \] and \[ y=5. \] Therefore, \[ x:y=6:5 \] Hence, \[ \boxed{6:5} \]
ion-electron method (also known as the oxidation number method). The key idea is that the total number of electrons lost during oxidation must be equal to the total number of electrons gained during reduction. We first identify the oxidation and reduction half-reactions and then combine them appropriately.
Step 1: Identify the species undergoing oxidation and reduction. The reactants are: \[ MnO_4^-, \quad I^-, \quad H^+ \] The products are: \[ Mn^{2+}, \quad IO_3^-, \quad H_2O \] For manganese: \[ MnO_4^- \] Oxidation state of Mn: \[ x+4(-2)=-1 \] \[ x-8=-1 \] \[ x=+7 \] In the product: \[ Mn^{2+} \] Oxidation state = \(+2\). Thus manganese undergoes reduction: \[ Mn^{+7}\rightarrow Mn^{+2} \] Gain in electrons: \[ 5e^- \] Therefore, each permanganate ion gains \(5\) electrons.
Step 2: Determine oxidation of iodide ion. For iodide: \[ I^- \rightarrow IO_3^- \] Oxidation state of iodine in \(I^-\): \[ -1 \] Oxidation state of iodine in \(IO_3^-\): \[ x+3(-2)=-1 \] \[ x-6=-1 \] \[ x=+5 \] Thus iodine changes from \[ -1 \rightarrow +5 \] Increase in oxidation number: \[ 6 \] Hence each iodide ion loses \[ 6e^- \] during oxidation.
Step 3: Write the half-reactions. Reduction half-reaction: \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+}+4H_2O \] Oxidation half-reaction: \[ I^-+3H_2O \rightarrow IO_3^-+6H^++6e^- \]
Step 4: Equalize the electrons exchanged. The reduction half-reaction involves \[ 5e^- \] while the oxidation half-reaction involves \[ 6e^-. \] The least common multiple of \(5\) and \(6\) is \[ 30. \] Therefore, Multiply the reduction half-reaction by \(6\): \[ 6MnO_4^- +48H^+ +30e^- \rightarrow 6Mn^{2+}+24H_2O \] Multiply the oxidation half-reaction by \(5\): \[ 5I^-+15H_2O \rightarrow 5IO_3^-+30H^++30e^- \]
Step 5: Add the two balanced half-reactions. Adding and cancelling electrons: \[ 6MnO_4^-+5I^-+18H^+ \rightarrow 6Mn^{2+}+5IO_3^-+9H_2O \] Thus the balanced equation is: \[ 6MnO_4^-+5I^-+18H^+ \rightarrow 6Mn^{2+}+5IO_3^-+9H_2O \]
Step 6: Determine the required ratio. Comparing with \[ xMnO_4^- + yI^- + zH^+ \rightarrow pMn^{2+}+qIO_3^-+rH_2O \] we obtain: \[ x=6 \] and \[ y=5. \] Therefore, \[ x:y=6:5 \] Hence, \[ \boxed{6:5} \]
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