Question

Use the oxidation number method to balance these equations.

a. H2SO4 + HI → S + I2 + H2O

b. HBr + H2SO4 → Br2 + SO2 + H2O

c. V3+ + I2 + H2O →  VO2+ + I– + H+

Answer 1

a) To balance a chemical reaction via oxidation method provided the should occur oxidation and reduction(I.e redox reaction).

•Given reaction:

H2SO4+ HI =S +I2 +H2O.

•Detection of oxidation no.

H2SO4,

+1×2 +y +(-2)4=0

Or, 2+y-8=0

Or, y= +6             so,  S in H2SO4 is +6

 

I, in HI is -1 (+1 +y =0 or, y=-1)

S and I2 are in atomic and molecular state in resultant.So their oxidation no. will be 0(zero).

 

•Redox explanation:

S in H2SO4 ,the oxidation no. of S decreases from +6 to 0 (6 units)  in S in the resultant.So reduction occurs. 

On the other hand,

I in HI is oxidized to I2, as the oxidation no enhanced from (-1to 0). Oxidation occurs. 

As I2(iodine molecule)  is diatonic so 3 is multiplied  to I2 ,and 6 will be multiplied to HI.

Now the equation  becomes 

 

H2SO4+6HI =S+3I2 +H2O

After this redox multiplication O and H atom are not balanced. So ,to balance it4 should be multiplied to H2O i.e 4H2O.

▪︎▪Final balanced equation:

H2SO4 + 6HI = S + 3 I2 + 4H2O

 

Reactant atoms:[H=2+6=8,S=1,O=4, I =6]

Product atoms,

[S=1,I=3×2=6,H=4×2=8,O=4]

SO BALANCED. 

b)

•Reaction:

 HBr + H2SO4 =Br2 + SO2 + H2O

•Detection of oxidation no.

Br in HBr =-1 (+1+x=0 ;x=-1)

S in H2SO4 =+6 (as before)

S in SO2 =+4 (y+(-2×2)=0, y=+4)

Br2 is 0.

 

So, Br in HBr oxidised to Br2 as oxidation no enhanced -1to 0. Oxidation occurs. 

Other way,the oxidation no of S in H2SO4 is decreases from +6 to +4 in SO2 gas. So decreases by 2 units.i.e reduction occurs. 

•Therefore the reaction becomes:

2HBr + H2SO4 =Br2 +SO2 +2H2O  (balanced)

 

c)

As the given reaction is not a redox reaction it is to be balanced by algebraic method. 

•Step 1: The equation is labelled by 

aV^+3 +bI2 + cH2O= d VO2 + f I + 8H^+

Step 2:

By using trial and error method,if V3^+ is multiplied by 2 then (6^+) charge on V , so in product  2 should be multiplied(2VO2) .After multiplying4 O atom in the product.so 4 is multiplied to H2O (4H2O).Thus 8 H atoms will be in the reactant,then 8 should be multiplied in H^+ ion(8H^+) . Since I2 in the reactant ,then in resultant 2 will be multiplied  in I^- (2I^-)

3rd step:

Simplifying the result to get the lowest whole integer value.

a=2{V^3+}

b=1(I2)

c=4(H2O)

d=2(VO2)

f=2(I^-)

g=8(H^+)

 

4th step:

▪︎▪Final balanced  reaction:

2 V^(3+) +I2 +4H2O= 2VO2 +2I^(-)  + 8 H^+.

----    ---    ------         ---   -------  --

g=8(H^+)

 

4th step:

▪︎▪Final balanced  reaction:

2 V^(3+) +I2 +4H2O= 2VO2 +2I^(-)  + 8 H^+.

----    ---    ------         ---   -------  --