Question

Balance the following equation and find the values of a, b, c and d.
$aMn{O}_4^{\ominus }+b{Br}^{\ominus }⟶c{Mn}^{2+}+d{Br}_2$ (in acid solution)

• A. $a=2,b=14,c=2,d=7$
• B. $a=2,b=10,c=2,d=5$
• C. $a=1,b=5,c=1,d=5$
• D. None of these

Answer 1

The correct option is B $a=2,b=10,c=2,d=5$

The unbalanced redox equation is as follows:
$Mn{O}_4^{\ominus }+{Br}^{\ominus }⟶{Mn}^{2+}+{Br}_2$
Balance all atoms other than H and O.
$Mn{O}_4^{\ominus }+2{Br}^{\ominus }⟶{Mn}^{2+}+{Br}_2$
The oxidation number of Mn changes from 7 to 2. The change in the oxidation number of Mn is 5.
The oxidation number of Br changes from -1 to 0. The change in the oxidation number of one Br atom is 1.
For two Br atoms, the oxidation number changes by 2.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying
$Mn{O}_4^{\ominus }$ and ${Mn}^{2+}$ with 2 and by multiplying ${Br}^{\ominus }$ and ${Br}_2$ with 5.
$2Mn{O}_4^{\ominus }+10{Br}^{\ominus }⟶2{Mn}^{2+}+5{Br}_2$
O atoms are balanced by adding 8 water molecules on RHS.
$2Mn{O}_4^{\ominus }+10{Br}^{\ominus }⟶2{Mn}^{2+}+5{Br}_2+8H_{2}O$
Hydrogen atoms are balanced by adding 16 $H^{+}$ atoms on LHS..
$2Mn{O}_4^{\ominus }+10{Br}^{\ominus }⟶2{Mn}^{2+}+5{Br}_2+8H_{2}O$
This is the balanced chemical equation.
Hence, $a=2,b=10,c=2,d=5$