First, split the above equation in two parts and then balance
individually to get balanced half-reactions:
Cl2 (aq) ---------> Cl-
(aq)
------(1) and
S2O32-(aq) ----------> SO42-
(aq)
-------(2)
First, balance eq (1) step by step as:
Cl2 (aq) ---------> Cl- (aq)
Cl2 (aq) ---------> 2Cl-
(aq) (balance Cl)
Now, balance charge to get balanced first half reaction as:
Cl2 (aq) + 2e-
---------> 2Cl- (aq)
-------------(3)
Now, balance second half reaction (2) as:
S2O32-(aq) ----------> SO42-
(aq)
S2O32-(aq) ---------->
2SO42-
(aq)
(balance S)
S2O32-(aq) + 5H2O
----------> 2SO42-
(aq) (balance
O)
S2O32-(aq) + 5H2O
----------> 2SO42- (aq)
+ 10H+ (balance H)
Now, balance charge to get balaced second half reaction as:
S2O32-(aq) + 5H2O
----------> 2SO42- (aq)
+ 10H+ + 4e-
---------(4)
Now, to balance charge between eq (3) and (4) multiply eq (3) by 2
as:
2Cl2 (aq) + 4e-
---------> 4Cl- (aq)
-----------(5)
Now, add eq (4) and (5) to get balanced reaction:
2Cl2 (aq) + 4e- + S2O32-(aq)
+ 5H2O ---------> 4Cl-
(aq) + 2SO42- (aq)
+ 10H+ + 4e-
Remove common parts from both sides to get balanced reaction
as:
2Cl2 (aq) + S2O32-(aq) +
5H2O ---------> 4Cl- (aq)
+ 2SO42- (aq)
+ 10H+
Correct option: (a) 10H+
QUESTION 1 Balance the following equation in acidic solution using the lowest possible integers a...
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