Question

tutor balance redox reactions in basic solution
balance the reaction between cl⁻ and h₂bo₃⁻ to form cl₂ and b in basic solution.
when you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown. enter \1\ if the coefficient is \1.\
6 cl⁻+ h₂bo₃⁻→3 cl₂+ b
water appears in the balanced equation as a reactant (reactant, product, neither) with a coefficient of . (enter 0 for neither.)
how many electrons are transferred in this reaction? 6
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Explanation:

Step1: Write half - reactions

Oxidation: $2Cl^-
ightarrow Cl_2 + 2e^-$
Reduction: $H_2BO_3^-
ightarrow B$

Step2: Balance non - H and non - O elements in reduction half - reaction

$H_2BO_3^-
ightarrow B$ is already balanced for B.

Step3: Balance O in reduction half - reaction in basic solution

$H_2BO_3^-
ightarrow B+3OH^-$

Step4: Balance H in reduction half - reaction in basic solution

$H_2BO_3^- + 4H_2O
ightarrow B+3OH^-$

Step5: Balance charge in reduction half - reaction

$H_2BO_3^- + 4H_2O+3e^-
ightarrow B+3OH^-$

Step6: Make electrons equal in both half - reactions

Multiply oxidation half - reaction by 3 and reduction half - reaction by 2.
Oxidation: $6Cl^-
ightarrow 3Cl_2 + 6e^-$
Reduction: $2H_2BO_3^- + 8H_2O+6e^-
ightarrow 2B+6OH^-$

Step7: Add half - reactions

$6Cl^-+2H_2BO_3^- + 8H_2O
ightarrow 3Cl_2+2B+6OH^-$

Answer:

$6Cl^- + 2H_2BO_3^-
ightarrow 3Cl_2+2B$
Water appears in the balanced equation as a reactant with a coefficient of 8.
The number of electrons transferred is 6.