1
Question
In the reaction,
xOH−+yAs2S3+zClO−3⟶aAsO3−4+bClO−+3SO2−4+cH2O,
x,y,z,a,b, and c are respectively:
In the reaction,
xOH−+yAs2S3+zClO−3⟶aAsO3−4+bClO−+3SO2−4+cH2O,
x,y,z,a,b, and c are respectively:
Open in App
Solution
The correct option is D 24,2,14,4,14,12
a) First, let us split As2S3 into As3+ and S2− ions where As3+ is oxidized to AsO3−4 while S2− is oxidized to SO2−4. Whereas ClO−3 is reduced to ClO−.
b) Let us write balance the electrons for As3+⟶AsO3−4 as
2+3As3+⟶2+5AsO3−4+4e−
c) Now let us balance the O atoms in the above reaction by adding H2O
2+3As3++8H2O⟶2+5AsO3−4+4e−
As you can see, there is an excess of 16H atoms on the LHS. To balance this, we add OH− to the LHS (since the medium is basic) and also add an equal number of H2O to the RHS. The reaction now looks like,
16OH−+2As3++8H2O⟶2AsO3−4+4e−+16H2O
which can be simplified as,
16OH−+As6+2+⟶2AsO3−4+4e−+8H2O ...(i)
d) Now let us balance the other half oxidation reaction S2−⟶SO2−4 as
S6−3⟶3SO2−4+24e−
e) in the above reaction, let us add 12H2O to the LHS to balance the O atoms
12H2O+S6−3⟶3SO2−4+24e−
In the above reaction, as you can see, there is an excess of 24H atoms on the LHS. To balance this, we add OH− to the LHS (since the medium is basic) and also add an equal number of H2O to the RHS. The reaction now looks like,
24OH−+12H2O+S6−3⟶3SO2−4+24e−+24H2O
which can be simplified to
24OH−+S6−3⟶3SO2−4+24e−+12H2O ...(ii)
Now, by adding the two half oxidation reactions (i) and (ii), we get the overall half oxidation reaction as
40OH−+As2S3⟶2AsO3−4+3SO2−4+28e−+20H2O ...(iii)
f) Balancing the only reduction half reaction takes a similar approach to the above two.
First, let us balance the electrons in ClO−3⟶ClO−
4e−+ClO−3⟶ClO−
In the above reaction, let us balance for the O atoms by adding \2H+2O\) to the RHS.
4e−+ClO−3⟶ClO−+2H2O
g) In the above reaction we can see that there is an excess of 4H atoms on the RHS. So we add 4OH− to the RHS
and also add 4H2O to the LHS. The reaction now looks like,
4e−+ClO−3+4H2O⟶ClO−+2H2O+4OH−
which can be simplified to
4e−+ClO−3+2H2O⟶ClO−+4OH− ...(iv)
h) multiplying equation (iii) × 2 && equation (iv) × 14 and adding, we get the balanced overall redoc reaction as
24OH−+2As2S3+14ClO−3⟶4AsO3−4+14ClO−+12H2O+6SO2−4
Hence x = 24, y = 2, z = 14, a = 4, b = 14, c = 12.
a) First, let us split As2S3 into As3+ and S2− ions where As3+ is oxidized to AsO3−4 while S2− is oxidized to SO2−4. Whereas ClO−3 is reduced to ClO−.
b) Let us write balance the electrons for As3+⟶AsO3−4 as
2+3As3+⟶2+5AsO3−4+4e−
c) Now let us balance the O atoms in the above reaction by adding H2O
2+3As3++8H2O⟶2+5AsO3−4+4e−
As you can see, there is an excess of 16H atoms on the LHS. To balance this, we add OH− to the LHS (since the medium is basic) and also add an equal number of H2O to the RHS. The reaction now looks like,
16OH−+2As3++8H2O⟶2AsO3−4+4e−+16H2O
which can be simplified as,
16OH−+As6+2+⟶2AsO3−4+4e−+8H2O ...(i)
d) Now let us balance the other half oxidation reaction S2−⟶SO2−4 as
S6−3⟶3SO2−4+24e−
e) in the above reaction, let us add 12H2O to the LHS to balance the O atoms
12H2O+S6−3⟶3SO2−4+24e−
In the above reaction, as you can see, there is an excess of 24H atoms on the LHS. To balance this, we add OH− to the LHS (since the medium is basic) and also add an equal number of H2O to the RHS. The reaction now looks like,
24OH−+12H2O+S6−3⟶3SO2−4+24e−+24H2O
which can be simplified to
24OH−+S6−3⟶3SO2−4+24e−+12H2O ...(ii)
Now, by adding the two half oxidation reactions (i) and (ii), we get the overall half oxidation reaction as
40OH−+As2S3⟶2AsO3−4+3SO2−4+28e−+20H2O ...(iii)
f) Balancing the only reduction half reaction takes a similar approach to the above two.
First, let us balance the electrons in ClO−3⟶ClO−
4e−+ClO−3⟶ClO−
In the above reaction, let us balance for the O atoms by adding \2H+2O\) to the RHS.
4e−+ClO−3⟶ClO−+2H2O
g) In the above reaction we can see that there is an excess of 4H atoms on the RHS. So we add 4OH− to the RHS
and also add 4H2O to the LHS. The reaction now looks like,
4e−+ClO−3+4H2O⟶ClO−+2H2O+4OH−
which can be simplified to
4e−+ClO−3+2H2O⟶ClO−+4OH− ...(iv)
h) multiplying equation (iii) × 2 && equation (iv) × 14 and adding, we get the balanced overall redoc reaction as
24OH−+2As2S3+14ClO−3⟶4AsO3−4+14ClO−+12H2O+6SO2−4
Hence x = 24, y = 2, z = 14, a = 4, b = 14, c = 12.
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
