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Question

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

$x M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) \to y M n O_{2} (s) + 2 H_{2} O + z O H^{-}$

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

A

x = 5, y = 3, z = 2

B

x = 3, y = 5, z = 2

C

x = 3, y = 3, z = 2

D

x = 3, y = 5, z = 3

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Solution

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The unbalanced chemical equation is as shown.

$M n {(O H)}_{2} (s) + M n O {_{4}}^{-} (a q) \to M n O_{2} (s)$

Balance $M n$ atoms.

$M n {(O H)}_{2} (s) + M n O {_{4}}^{-} (a q) \to 2 M n O_{2} (s)$

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

The oxidation number of $M n$ 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.

$3 M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) \to 5 M n O_{2} (s)$

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

$3 M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) \to 5 M n O_{2} (s) + 4 H_{2} O (a q)$

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

$3 M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) + 2 H^{+} (a q) \to 5 M n O_{2} (s) + 4 H_{2} O (a q)$

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

$3 M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) + 2 H^{+} (a q) + 2 O H^{-} (a q) \to 5 M n O_{2} (s) + 4 H_{2} O (a q) + 2 O H^{-} (a q)$

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

$3 M n {(O H)}_{2} (s) + 2 M n O {_{4}}^{-} (a q) \to 5 M n O_{2} (s) + 2 H_{2} O (a q) + 2 O H^{-} (a q)$

This is the balanced chemical equation.

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

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