Question

In the reaction,
$xO{H}^{-}+yA{s}_2{S}_3+zCl{O}_3^{-}⟶aAs{O}_4^{3-}+bCl{O}^{-}+3S{O}_4^{2-}+c{H}_{2}O$,

x,y,z,a,b, and c are respectively:

• A. 2,14,24,14,4,12
• B. 24,4,14,2,14,12
• C. 20,4,12,4,12,10
• D. 24,2,14,4,14,12

Answer 1

The correct option is D 24,2,14,4,14,12

a) First, let us split $A{s}_2{S}_3$ into $A{s}^{3+}$ and $S^{2-}$ ions where $A{s}^{3+}$ is oxidized to $As{O}_4^{3-}$ while $S^{2-}$ is oxidized to $S{O}_4^{2-}$. Whereas $Cl{O}_3^{-}$ is reduced to $Cl{O}^{-}$.

b) Let us write balance the electrons for $A{s}^{3+}⟶As{O}_4^{3-}$ as
$2\stackrel{+3}{A}{s}^{3+}⟶2\stackrel{+5}{A}s{O}_4^{3-}+4e^{-}$

c) Now let us balance the O atoms in the above reaction by adding $H_{2}O$
$2\stackrel{+3}{A}{s}^{3+}+8H_{2}O⟶2\stackrel{+5}{A}s{O}_4^{3-}+4e^{-}$

As you can see, there is an excess of 16H atoms on the LHS. To balance this, we add $O{H}^{-}$ to the LHS (since the medium is basic) and also add an equal number of $H_{2}O$ to the RHS. The reaction now looks like,

$16O{H}^{-}+2A{s}^{3+}+8H_{2}O⟶2As{O}_4^{3-}+4e^{-}+16H_{2}O$

which can be simplified as,
$16O{H}^{-}+A{s}_2^{6+}+⟶2As{O}_4^{3-}+4e^{-}+8H_{2}O$ ...(i)

d) Now let us balance the other half oxidation reaction $S^{2-}⟶S{O}_4^{2-}$ as

$S_3^{6-}⟶3S{O}_4^{2-}+24e^{-}$

e) in the above reaction, let us add $12H_{2}O$ to the LHS to balance the O atoms

$12H_{2}O+S_3^{6-}⟶3S{O}_4^{2-}+24e^{-}$

In the above reaction, as you can see, there is an excess of 24H atoms on the LHS. To balance this, we add $O{H}^{-}$ to the LHS (since the medium is basic) and also add an equal number of $H_{2}O$ to the RHS. The reaction now looks like,

$24O{H}^{-}+12H_{2}O+S_3^{6-}⟶3S{O}_4^{2-}+24e^{-}+24H_{2}O$

which can be simplified to
$24O{H}^{-}+S_3^{6-}⟶3S{O}_4^{2-}+24e^{-}+12H_{2}O$ ...(ii)

Now, by adding the two half oxidation reactions (i) and (ii), we get the overall half oxidation reaction as

$40O{H}^{-}+A{s}_2{S}_3⟶2As{O}_4^{3-}+3S{O}_4^{2-}+28e^{-}+20H_{2}O$ ...(iii)

f) Balancing the only reduction half reaction takes a similar approach to the above two.

First, let us balance the electrons in $Cl{O}_3^{-}⟶Cl{O}^{-}$ ​

$4e^{-}+Cl{O}_3^{-}⟶Cl{O}^{-}$ ​

In the above reaction, let us balance for the O atoms by adding \2H+2O\) to the RHS.

$4e^{-}+Cl{O}_3^{-}⟶Cl{O}^{-}+2H_{2}O$

g) In the above reaction we can see that there is an excess of 4H atoms on the RHS. So we add $4O{H}^{-}$ to the RHS
and also add $4H_{2}O$ to the LHS. The reaction now looks like,

$4e^{-}+Cl{O}_3^{-}+4H_{2}O⟶Cl{O}^{-}+2H_{2}O+4O{H}^{-}$

which can be simplified to ​

$4e^{-}+Cl{O}_3^{-}+2H_{2}O⟶Cl{O}^{-}+4O{H}^{-}$ ...(iv)


h) multiplying equation (iii) $\times$ 2 && equation (iv) $\times$ 14 and adding, we get the balanced overall redoc reaction as

$24O{H}^{-}+2A{s}_2{S}_3+14Cl{O}_3^{-}⟶4As{O}_4^{3-}+14Cl{O}^{-}+12H_{2}O+6S{O}_4^{2-}$

Hence x = 24, y = 2, z = 14, a = 4, b = 14, c = 12.