Question

$A{s}_2{S}_3+xN{O}_3^{-}+y{H}^{+}\to 2As{O}_4^{3-}+zN{O}_2+aS+2H_{2}O$ Here $x,y,z$ and $a$ respectively are

• A. 10, 4, 10, 3
• B. 5, 4, 5, 3
• C. 4, 10, 3, 10
• D. None of these

Answer 1

The correct option is A 10, 4, 10, 3

This is a very special case of redox reactions. Note that the given reaction involves two oxidations and one reduction process.
a. Two oxidation half reactions are
$i.A{s}^{3+}\to As{O}_4^{3-}ii.S^{2-}\to S$
Now balance and add the two equations to get
$2A{s}^{3+}+3S^{2-}+8H_{2}O\to 2As{O}_4^{3-}+3S+16H^{+}+10e^{-}$
b. Reduction half reaction: $N{O}_3^{-}\to N{O}_2$
Now balance the reduction half reaction and proceed as usual .
$N{O}_3^{-}+2H^{+}+1e^{-}\to N{O}_2+H_{2}O$
Adding the two reactions (i) and (ii), we get
$A{s}_2{S}_3+10N{O}_3^{-}+4H^{+}\to 2As{O}_4^{3-}+10N{O}_2+3S+2H_{2}O$
Note : In the above type of cases, take the total change in oxidation state (increase or decrease) for cross multiplication while adding oxidation and reduction half reactions. For example, in the given case, 10 electrons are lost per mole of $A{s}_2{S}_3$.