Question

Consider the following reaction:

$xMnO_4^- + yI^- + zH^+ \longrightarrow pMn^{2+} + qIO_3^- + rH_2O$

The correct ratio $x:y$ in the balanced equation is

• A. 6:5
• B. 6:4
• C. 1:1
• D. 5:4

Answer 1

Answer: (A)

6:5

Answer 2

The given reaction is:

$xMnO_4^- + yI^- + zH^+ \longrightarrow pMn^{2+} + qIO_3^- + rH_2O$

We will balance this redox reaction using the ion-electron method in acidic medium.

Step 1: Identify oxidation and reduction half-reactions.

• In $MnO_4^-$, Mn is in the +7 oxidation state. In $Mn^{2+}$, Mn is in the +2 oxidation state. Thus, $MnO_4^-$ is reduced.
• In $I^-$, I is in the -1 oxidation state. In $IO_3^-$, let the oxidation state of I be 'a'. Since oxygen is -2, we have a + 3(-2) = -1, which gives a - 6 = -1, so a = +5. Thus, $I^-$ is oxidized.

Reduction half-reaction: $MnO_4^- \longrightarrow Mn^{2+}$ Oxidation half-reaction: $I^- \longrightarrow IO_3^-$

Step 2: Balance atoms other than O and H.

• Manganese atoms are balanced.
• Iodine atoms are balanced.

Step 3: Balance oxygen atoms by adding $H_2O$.

• Reduction: $MnO_4^- \longrightarrow Mn^{2+} + 4H_2O$ (4 oxygen atoms on the left, so add 4 $H_2O$ to the right)
• Oxidation: $I^- + 3H_2O \longrightarrow IO_3^-$ (3 oxygen atoms on the right, so add 3 $H_2O$ to the left)

Step 4: Balance hydrogen atoms by adding $H^+$ (since it's an acidic medium).

• Reduction: $MnO_4^- + 8H^+ \longrightarrow Mn^{2+} + 4H_2O$ (8 hydrogen atoms on the right from $4H_2O$, so add 8 $H^+$ to the left)
• Oxidation: $I^- + 3H_2O \longrightarrow IO_3^- + 6H^+$ (6 hydrogen atoms on the left from $3H_2O$, so add 6 $H^+$ to the right)

Step 5: Balance charge by adding electrons ($e^-$).

• Reduction: $MnO_4^- + 8H^+ \longrightarrow Mn^{2+} + 4H_2O$
  • Charge on left: (-1) + (+8) = +7
  • Charge on right: +2
  • To balance, add 5 electrons to the left side: $MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$
• Oxidation: $I^- + 3H_2O \longrightarrow IO_3^- + 6H^+$
  • Charge on left: -1
  • Charge on right: (-1) + (+6) = +5
  • To balance, add 6 electrons to the right side: $I^- + 3H_2O \longrightarrow IO_3^- + 6H^+ + 6e^-$

Step 6: Equalize the number of electrons transferred in both half-reactions.

The least common multiple of 5 and 6 is 30.

• Multiply the reduction half-reaction by 6: $6(MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O)$ $6MnO_4^- + 48H^+ + 30e^- \longrightarrow 6Mn^{2+} + 24H_2O$
• Multiply the oxidation half-reaction by 5: $5(I^- + 3H_2O \longrightarrow IO_3^- + 6H^+ + 6e^-)$ $5I^- + 15H_2O \longrightarrow 5IO_3^- + 30H^+ + 30e^-$

Step 7: Add the balanced half-reactions and cancel common species.

$(6MnO_4^- + 48H^+ + 30e^- \longrightarrow 6Mn^{2+} + 24H_2O)$ $+$ $(5I^- + 15H_2O \longrightarrow 5IO_3^- + 30H^+ + 30e^-)$

$6MnO_4^- + 48H^+ + 30e^- + 5I^- + 15H_2O \longrightarrow 6Mn^{2+} + 24H_2O + 5IO_3^- + 30H^+ + 30e^-$

Cancel $30e^-$ from both sides. Cancel $30H^+$ from $48H^+$ on the left, leaving $18H^+$ on the left. Cancel $15H_2O$ from $24H_2O$ on the right, leaving $9H_2O$ on the right.

The balanced equation is:

$6MnO_4^- + 5I^- + 18H^+ \longrightarrow 6Mn^{2+} + 5IO_3^- + 9H_2O$

Step 8: Determine the ratio $x:y$.

From the balanced equation, $x = 6$ and $y = 5$. The ratio $x:y$ is $6:5$.