✔ 最佳答案
As there are a number of acids to learn in chemistry, you should know how to deduce them.
First, bromine water is an oxidizing agent. It would be decolourised when reacting with reducing agent. Therefore, you should think about a gas that satisfied the above situation.
The answer will then come to your mind immediately. That is concentrated sulphuric acid. Concentrated sulphuric acid is not acting as an acid in this case, it will not give hydrogen gas but sulphur dioxide gas due to its oxidizing properties.
2H2SO4 (l) + Zn (s) ---> ZnSO4 (aq) + 2SO2 (g) + 2H2O (l)
SO2 formed is the reducing agent.
Then, you can start to write the ionic-half equation.
SO2 (g) + 2H2O (l) ---> SO4 2- (aq) + 4H+ (aq) + 2e- --------------(1)
Br2 (aq) + 2e- ---> 2Br- (aq) ----------(2)
Since both equation have the same charge, you can simply combine them.
The combined equation is as follows: (no need to write the electrons anymore as they will be cancelled at both sides together finally)
SO2 (g) + 2H2O (l) + Br2 (l) --> SO4 2- (aq) + 4H+ (aq) + 2Br- (aq)
Note: writing ionic-half equation & the combined equation afterwards should include STATE SYMBOLS.
Hope my explanations help.
參考: myself.