Explain how to balance redox reactions with an example
Answers
Answer:
Method: Oxidation number method
Redox reaction is balanced in two mediums
1) In Acidic medium:
e.g. SO2 (g) + Fe³+ (aq) →
Fe²+ (aq) + (SO4)²- (aq)
Oxidation number in L.H.S.
→ S = 4, Fe = 3
Oxidation number in R.H.S.
→ S = 6, Fe = 2
Note: Don't consider oxygen.
For S, net increase = 2
For Fe, net decrease = - 1
Here, net » for no. of same atoms present.
Step 1:
Multiply Fe by 2 to balance
net increase = magnitude of net decrease
» SO2 (g) + 2Fe³+ (aq) →
2Fe²+ (aq) + (SO4)²- (aq)
Step 2:
Balancing O (oxygen) atoms
In, L.H.S. = 2 & R.H.S. = 4
Add nH2O to side where there is less no. of O. n is the number by which it is less.
» SO2 (g) + 2Fe³+ (aq) + 2H2O (l) →
2Fe²+ (aq) + (SO4)²- (aq)
Step 3:
Balance H (hydrogen) atoms
In, L.H.S. = 4 & R.H.S. = 0
Add nH+ ion to side where there is less no. of H. n is the number by which it is less.
» SO2 (aq) + 2Fe³+ (aq) + 2H2O (l) →
2Fe²+ (aq) + (SO4)²- + 4H+ (aq)
Above reaction is balanced.
______________________________
2) Basic medium:
MnCl2 (aq) + H2O2 (aq) →
Mn (OH)3 (aq) + 2Cl- (aq)
Similar till Step 3:
» 2MnCl2 (aq) + H2O2 (aq) + 4H2O (l) →
2Mn(OH)3 + 4Cl- (aq) + 4H+ (aq)
Step 4:
Add OH- on both side as same no. H+ added.
» 2MnCl2 (aq) + H2O2 (aq) + 4H2O (l) + 4OH- →
2Mn(OH)3 + 4Cl- (aq) + 4H+ (aq) + 4OH-
Step 5:
Note: (H+) + OH- = H2O (l)
» 2MnCl2 (aq) + H2O2 (aq) + 4H2O (l) + 4OH- →
2Mn(OH)3 + 4Cl- (aq) + 4H2O (l)
Above reaction is balanced.
Answer:
A redox equation can be balanced using the following stepwise procedure:
(1) Divide the equation into two half-reactions.
(2) Balance each half-reaction for mass and charge.
(3) Equalize the number of electrons transferred in each half-reaction.
(4) Add the half-reactions together.
Explanation:
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