Question

$M{n}^{3+}$ ions are unstable in solution and undergo disproportionation to give $M{n}^{2+}$, $Mn{O}_2$ and H^+ ions. What will be the balanced equation for the reaction?

• A. $3M{n}^{3+}+4H_{2}O\to Mn{O}_2+M{n}^{2+}+8H^{+}$
• B. $M{n}^{3+}+4H_{2}O\to Mn{O}_2+4H^{+}$
• C. $Mn+2H_{2}O\to Mn{O}_2+4H^{+}$
• D. $2M{n}^{3+}+2H_{2}O\to Mn{O}_2+M{n}^{2+}+4H^{+}$

Answer 1

The correct option is D $2M{n}^{3+}+2H_{2}O\to Mn{O}_2+M{n}^{2+}+4H^{+}$

In a balanced chemical reaction number of all atoms at right side should be equal to the left side of the reaction.

Given reaction: $M{n}^{3+}+H_{2}O\to Mn{O}_2+M{n}^{2+}+H^{+}$

Balancing a chemical reaction as:

Step 1: Assign oxidation numbers to each of the atoms in the equation and write the numbers above the atom as:

$\stackrel{+3}{M{n}^{3+}}+H_{2}O\to \stackrel{+4}{Mn}{O}_2+\stackrel{+2}{M{n}^{2+}}+H^{+}$

Step 2: Identify the atoms that are oxidized and those that are reduced as:

Reduction: $\stackrel{+3}{M{n}^{3+}}\to \stackrel{+2}{M{n}^{2+}}$

Oxidation: $\stackrel{+3}{M{n}^{3+}}\to \stackrel{+4}{Mn}{O}_2$

Step 3: oxidation-number change is:

Reduction: $\stackrel{+3}{M{n}^{3+}}\to \stackrel{+2}{M{n}^{2+}}$: gain of total 1 electrons

Oxidation: $\stackrel{+3}{M{n}^{3+}}\to \stackrel{+4}{Mn}{O}_2$: Loss of total 1 electrons

Step 4: The total change in oxidation number is balanced.

Step 5: Balance O atoms in oxidation reaction by adding $H_{2}O$ and then balance H by $H^{+}$ as:

Oxidation: $\stackrel{+3}{M{n}^{3+}}+2H_{2}O\to \stackrel{+4}{MnS}{O}_2+4H^{+}$

Thus overall Balanced reaction is:

$\stackrel{+3}{M{n}^{3+}}+\stackrel{+3}{M{n}^{3+}}+2H_{2}O\to \stackrel{+2}{M{n}^{2+}}+\stackrel{+4}{MnS}{O}_2+4H^{+}$

or $2M{n}^{3+}2H_{2}O\to M{n}^{2+}+Mn{O}_2+4H^{+}$

is the balanced reaction