Question

If the balance reaction is as given below, find the value of $x,y$ and $z$.

$xMn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)\to yMn{O}_2\left(s\right)+2H_{2}O+zO{H}^{-}$

• A. $x=3,y=5,z=2$
• B. $x=5,y=3,z=2$
• C. $x=3,y=3,z=2$
• D. $x=3,y=5,z=3$

Answer 1

Answer: (A)

$x=3,y=5,z=2$

The unbalanced chemical equation is as shown.

$Mn{\left(OH\right)}_2\left(s\right)+MnO{{}_4}^{-}\left(aq\right)\to Mn{O}_2\left(s\right)$

Balance $Mn$ atoms.

$Mn{\left(OH\right)}_2\left(s\right)+MnO{{}_4}^{-}\left(aq\right)\to 2Mn{O}_2\left(s\right)$

The oxidation number of $Mn$ increases from $+2$ (in manganese hydroxide) to $+4$ (in manganese dioxide). Increase in the oxidation number is $+2$.

The oxidation number of $Mn$ decreases from $+7$ (in permanganate ion) to $+4$ (in manganese dioxide). Decrease in the oxidation number is $+3$.

Balance the increase in the oxidation number with decrease in the oxidation number.

$3Mn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)\to 5Mn{O}_2\left(s\right)$

To balance $O$ atoms, add $4$ water molecules on RHS.

$3Mn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)\to 5Mn{O}_2\left(s\right)+4H_{2}O\left(aq\right)$

To balance $H$ atoms, add $2$ protons on LHS.

$3Mn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)+2H^{+}\left(aq\right)\to 5Mn{O}_2\left(s\right)+4H_{2}O\left(aq\right)$

Since, the reaction is in basic medium, add $2$ hydroxide ions on either side.

$3Mn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)+2H^{+}\left(aq\right)+2O{H}^{-}\left(aq\right)\to 5Mn{O}_2\left(s\right)+4H_{2}O\left(aq\right)+2O{H}^{-}\left(aq\right)$

On LHS, $2$ hydrogen ions combine with $2$ hydroxide ions to from $2$ water molecules which $2$ water molecules on RHS.

$3Mn{\left(OH\right)}_2\left(s\right)+2MnO{{}_4}^{-}\left(aq\right)\to 5Mn{O}_2\left(s\right)+2H_{2}O\left(aq\right)+2O{H}^{-}\left(aq\right)$

This is the balanced chemical equation.

The coefficients $x,y$ and $z$ are $3,5$ and $2$.