Question

Balance the following reaction under basic conditions. What are the coefficients in front of H2O and Cl- in the balanced reaction?

Cl2(aq) + Br2(l) → BrO3-(aq) + Cl-(aq)

Answer 1

`5Cl_2^0+Br_2^0+12OH^-\rightarrow 10Cl^-+2(Br^(+5)O^(-2)_3)^-+6H_2O`

Step-by-step explanation:

Hello,

In this case, for the given reaction, we first identify each element's oxidation state:

`Cl_2^0(aq) + Br_2^0(l) \rightarrow (Br^(+5)O_3^(-2))^-(aq) + Cl^-(aq)`

Hence, it seen that chlorine is reduced whereas bromine is oxidized, thereby, the half-reactions in acidic media (first step) are:

`Cl_2^0+2e^-\rightarrow 2Cl^-\\Br_2^0+6H_2O\rightarrow 2(Br^(+5)O^(-2)_3)^-+10e^-+12H^+`

Then we balance the electrons:

`10*(Cl_2^0+2e^-\rightarrow 2Cl^-)\\2*(Br_2^0+6H_2O\rightarrow 2(Br^(+5)O^(-2)_3)^-+10e^-+12H^+)`

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`10Cl_2^0+20e^-\rightarrow 20Cl^-\\2Br_2^0+12H_2O\rightarrow 4(Br^(+5)O^(-2)_3)^-+20e^-+24H^+\\`

Finally, we add as many OH⁻ as many H⁺ we have in order the obtain water and OH⁻ at each side once the half reactions are added:

`10Cl_2^0+2Br_2^0+12H_2O+24OH^-\rightarrow 20Cl^-+4(Br^(+5)O^(-2)_3)^-+24H^++24OH^-\\\\10Cl_2^0+2Br_2^0+12H_2O+24OH^-\rightarrow 20Cl^-+4(Br^(+5)O^(-2)_3)^-+24H_2O`

So the simplified result is:

`5Cl_2^0+Br_2^0+12OH^-\rightarrow 10Cl^-+2(Br^(+5)O^(-2)_3)^-+6H_2O`

Best regards.

Answer 2

3Br2 + 6OH- → 5Br- + BrO3- + 3H2O

Step-by-step explanation:

We first write the skeletal equation: Br2 → Br- + BrO3-

Next we obtain the oxidation and reduction half equation

Oxidation

Br2 + 12OH- → 2BrO3- 10e- + 6H2O

Reduction

5Br2 + 10e- → 10Br-

Next we put down the overall reaction equation simplifying the coefficients and cancelling electrons gained/lost:

3Br2 + 6OH- → 5Br- + BrO3- + 3H2O