Question

Balance the reaction between F₂ and Br₂ to form F^- and BrO₃^- in basic solution. When you have balanced the equation using the smallest integers possible, enter the coefficients of the species shown.

---------- F₂ + ------- Br_{2 =} --------- F^- + ---------- BrO₃^-

Water appears in the balanced equation as a  (reactant, product, neither) ----------- with a coefficient of ---------. (Enter 0 for neither.)

How many electrons are transferred in this reaction ---------- ?

Answer 1

F in F2 has oxidation state of 0

F in F- has oxidation state of -1

So, F in F2 is reduced to F-

Br in Br2 has oxidation state of 0

Br in BrO3- has oxidation state of +5

So, Br in Br2 is oxidised to BrO3-

Reduction half cell:

F2 + 2e- --> 2 F-

Oxidation half cell:

Br2 --> 2 BrO3- + 10e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

5 F2 + 10e- --> 10 F-

Oxidation half cell:

Br2 --> 2 BrO3- + 10e-

10 electrons are being transferred

Lets combine both the reactions.

5 F2 + Br2 --> 10 F- + 2 BrO3-

Balance Oxygen by adding water

5 F2 + Br2 + 6 H2O --> 10 F- + 2 BrO3-

Balance Hydrogen by adding H+

5 F2 + Br2 + 6 H2O --> 10 F- + 2 BrO3- + 12 H+

Add equal number of OH- on both sides as the number of H+

5 F2 + Br2 + 6 H2O + 12 OH- --> 10 F- + 2 BrO3- + 12 H+ + 12 OH-

Combine H+ and OH- to form water

5 F2 + Br2 + 6 H2O + 12 OH- --> 10 F- + 2 BrO3- + 12 H2O

Remove common H2O from both sides

Balanced Eqn is

5 F2 + Br2 + 12 OH- --> 10 F- + 2 BrO3- + 6 H2O

This is balanced chemical equation in basic medium

Coefficients are 5, 1, 10, 2

Water appears as product with coefficient of 6

10 electrons are being transferred