Question

The balanced equation in acidic medium is :

$S_2{O}_3^{2-}+S{b}_2{O}_5\stackrel{H^{+}}{\to }SbO+H_{2}S{O}_3$

• A. $3S_2{O}_3^{2-}+2S{b}_2{O}_5+6H^{+}3H_{2}O\to 4SbO+6H_{2}S{O}_3$
• B. $2S_2{O}_3^{2-}+2S{b}_2{O}_5+6H^{+}3H_{2}O\to 4SbO+4H_{2}S{O}_3$
• C. $3S_2{O}_3^{2-}+2S{b}_2{O}_5+4H^{+}2H_{2}O\to 4SbO+2H_{2}S{O}_3$
• D. none of these

Answer 1

The correct option is A $3S_2{O}_3^{2-}+2S{b}_2{O}_5+6H^{+}3H_{2}O\to 4SbO+6H_{2}S{O}_3$

The unbalanced redox equation is as follows:
$S_2{O}_3^{2-}+S{b}_2{O}_5\stackrel{H^{+}}{\to }SbO+H_{2}S{O}_3$
Balance all atoms other than H and O.
$S_2{O}_3^{2-}+S{b}_2{O}_5\stackrel{H^{+}}{\to }2SbO+2H_{2}S{O}_3$
The oxidation number of S changes from +2 to +4. The change in the oxidation number of S is 2. Total change in the oxidation number for 2 S atoms is 4.
The oxidation number of Sb changes from +5 to 3. The change in the oxidation number of 2 Sb is 6.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying $S_2{O}_3^{2-}$ and $H_{2}S{O}_3$ with 3 and multiplying $S{b}_2{O}_5$ and $SbO$ with 2.
$3S_2{O}_3^{2-}+2S{b}_2{O}_5\stackrel{H^{+}}{\to }4SbO+6H_{2}S{O}_3$
O atoms are balanced by adding 3 water molecules on LHS.
$3S_2{O}_3^{2-}+2S{b}_2{O}_5+3H_{2}O\stackrel{H^{+}}{\to }4SbO+6H_{2}S{O}_3$
Hydrogen atoms are balanced by adding 6 $H^{+}$ atoms on the LHS.
$3S_2{O}_3^{2-}+2S{b}_2{O}_5+3H_{2}O+6H^{+}\stackrel{H^{+}}{\to }4SbO+6H_{2}S{O}_3$
This is the balanced chemical equation.