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QUESTION 1 Balance the following equation in acidic solution using the lowest possible integers and give the coefficient of H

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Answer #1

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+

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