Write big and clearly please. QUESTION 1 Balance the following equation in acidic solution using the lowest possible integers and give the coefficient of H+ Cl 2(aq) + S 20 3 2-(aq)-Cl-(aq) + SO 4 2-(aq) a. 10 b. 2 ос.1 d. 8 e. 3 QUESTION 1 Balance the following equation in acidic solution using the lowest possible integers and give the coefficient of H+ Cl 2(aq) + S 20 3 2-(aq)-Cl-(aq) + SO 4 2-(aq) a. 10 b. 2 ос.1 d. 8 e. 3
If you split the equation in two parts, you can balance it 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) Balance charge is needed to get balanced first half reaction. 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) Balance charge to get a negative reaction in the second half. 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 in order to get balanced reaction. 2Cl2 (aq) + S2O32-(aq) + 5H2O ---------> 4Cl- (aq) + 2SO42- (aq) + 10H+ Correct option: (a) 10H+
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