Question

Consider the following redox equation Mn(OH)2(s) + MnO4 –(aq)  MnO42 –(aq) (basic solution) When the equation is balanced with smallest whole number coefficients, what is the coefficient for OH –(aq) and on which side of the equation is OH –(aq) present?

Answer 1

Answer : The balanced chemical equation is,

$2MnO_4^-(aq)+Mn(OH)_2(s)+4OH^-(aq)\rightarrow 3MnO_4^{2-}(aq)+2H_2O(l)$

The coefficient for $OH^-$ is, 4 and on reactant side of the equation $OH^-(aq)$ is present.

Explanation :

Rules for the balanced chemical equation in basic solution are :

• First we have to write into the two half-reactions.
• Now balance the main atoms in the reaction.
• Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
• If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
• If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.
• Now balance the charge.

The half reactions in the basic solution are :

Reduction : $Mn(OH)_2(s)+1e^-\rightarrow MnO_4^{2-}(aq)$ ......(1)

Oxidation : $MnO_4^-(aq)+4OH^-(aq)\rightarrow MnO_4^{2-}(aq)+2H_2O(l)+2e^-$.....(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2MnO_4^-(aq)+Mn(OH)_2(s)+4OH^-(aq)\rightarrow 3MnO_4^{2-}(aq)+2H_2O(l)$

Thus, the coefficient for $OH^-$ is, 4 and on reactant side of the equation $OH^-(aq)$ is present.