Write a balanced redox reaction between mercury and bromine to form Hg2^2+.
2Hg+Br2 >>Hg2++ + 2Br-
To write a balanced redox reaction between mercury (Hg) and bromine (Br2) to form Hg2^2+, we need to identify the oxidation states of the elements involved.
In the reactant side (left-hand side):
- Mercury is in its elemental state, which has an oxidation state of 0.
- Bromine is also in its elemental state, which also has an oxidation state of 0.
In the product side (right-hand side):
- Hg2^2+ has a net charge of 2+, so the oxidation state of mercury is +2.
- Bromine is present as an ion, so its oxidation state is -1.
To balance the redox reaction, we need to ensure that the total increase in oxidation state matches the total decrease in oxidation state.
Here's the balanced redox reaction:
Hg + Br2 → Hg2^2+ + 2Br^-
To balance the reaction:
1. Balance bromine atoms by placing a coefficient of 2 in front of HBr in the product side:
Hg + Br2 → Hg2^2+ + 2Br^-
2. Balance the charges by placing a coefficient of 2 in front of Hg in the reactant side:
2Hg + Br2 → Hg2^2+ + 2Br^-
This balanced redox reaction means that when 2 atoms of mercury react with 1 molecule of bromine, they form 1 ion of Hg2^2+ and 2 Br^- ions.