Question

In the redox reaction:
$xN{O}_3^{-}+yA{s}_2{S}_3+z{H}_{2}O\to As{O}_4^{3-}+NO+S{O}_4^{2-}+H^{+}$
The value of $\frac{x}{z}$ is:

Answer 1

1. Separate the reaction into half reactions. Identify and write out all redox couples in a reaction. Identify which reactant being oxidized and which being reduced.

O:${As}_2{S}_3\to 2As{O}_4^{3-}{As}_2{S}_3\to 3{SO}_4^{2-}$

R:${NO}_3^{-}\to NO$

2.Write down the transfer of electrons.

O: ${As}_2{S}_3\to 2As{O}_4^{3-}+4e^{-}{As}_2{S}_3\to 3{SO}_4^{2-}+24e^{-}$

R:${3e}^{-}+{NO}_3^{-}\to NO$

3.Combine these redox couples into two half reactions: one for oxidation and one for reduction

O:${As}_2{S}_3\to 2As{O}_4^{3-}+28e^{-}+3{SO}_4^{2-}$

R:${3e}^{-}+{NO}_3^{-}\to NO$

4.Balance the charge by adding $H^{+}$ ions to side deficient in positive charge.

O:${As}_2{S}_3\to 2As{O}_4^{3-}+28e^{-}+3{SO}_4^{2-}+40H^{+}$

R:$4H^{+}+{3e}^{-}+{NO}_3^{-}\to NO$

5. Balance the oxygen atoms by adding water molecules.

O:$20H_{2}O+{As}_2{S}_3\to 2As{O}_4^{3-}+28e^{-}+3{SO}_4^{2-}+40H^{+}$

R:$4H^{+}+{3e}^{-}+{NO}_3^{-}\to NO+2H_{2}O$

6.Make electron gain equivalent to electron lost.

O:$60H_{2}O+3{As}_2{S}_3\to 6As{O}_4^{3-}+84e^{-}+6{SO}_4^{2-}+120H^{+}$

R:$112H^{+}+{84e}^{-}+28{NO}_3^{-}\to 28NO+56H_{2}O$

7.Add the half reactions together, we get
$4H_{2}O+3{As}_2{S}_3+28{NO}_3^{-}\to 6As{O}_4^{3-}+28NO+6{SO}_4^{2-}+8H^{+}$

x=28 and z=4

So, $\frac{x}{z}$ is 7.