Answer:
3Br2 + 6OH- → 5Br- + BrO3- + 3H2O
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
Answer:
[tex]5Cl_2^0+Br_2^0+12OH^-\rightarrow 10Cl^-+2(Br^{+5}O^{-2}_3)^-+6H_2O[/tex]
Explanation:
Hello,
In this case, for the given reaction, we first identify each element's oxidation state:
[tex]Cl_2^0(aq) + Br_2^0(l) \rightarrow (Br^{+5}O_3^{-2})^-(aq) + Cl^-(aq)[/tex]
Hence, it seen that chlorine is reduced whereas bromine is oxidized, thereby, the half-reactions in acidic media (first step) are:
[tex]Cl_2^0+2e^-\rightarrow 2Cl^-\\Br_2^0+6H_2O\rightarrow 2(Br^{+5}O^{-2}_3)^-+10e^-+12H^+[/tex]
Then we balance the electrons:
[tex]10*(Cl_2^0+2e^-\rightarrow 2Cl^-)\\2*(Br_2^0+6H_2O\rightarrow 2(Br^{+5}O^{-2}_3)^-+10e^-+12H^+)[/tex]
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[tex]10Cl_2^0+20e^-\rightarrow 20Cl^-\\2Br_2^0+12H_2O\rightarrow 4(Br^{+5}O^{-2}_3)^-+20e^-+24H^+\\[/tex]
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:
[tex]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[/tex]
So the simplified result is:
[tex]5Cl_2^0+Br_2^0+12OH^-\rightarrow 10Cl^-+2(Br^{+5}O^{-2}_3)^-+6H_2O[/tex]
Best regards.
