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Balance the reaction between F2 and S2032 to form F and SO3 in basic solution. When you have balanced the equation using the

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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-

S in S2O3-2 has oxidation state of +2

S in SO3-2 has oxidation state of +4

So, S in S2O3-2 is oxidised to SO3-2

Reduction half cell:

F2 + 2e- --> 2 F-

Oxidation half cell:

S2O3-2 --> 2 SO3-2 + 4e-

Balance number of electrons to be same in both half reactions

Reduction half cell:

2 F2 + 4e- --> 4 F-

Oxidation half cell:

S2O3-2 --> 2 SO3-2 + 4e-

4 electrons are transferred

Lets combine both the reactions.

2 F2 + S2O3-2 --> 4 F- + 2 SO3-2

Balance Oxygen by adding water

2 F2 + S2O3-2 + 3 H2O --> 4 F- + 2 SO3-2

Balance Hydrogen by adding H+

2 F2 + S2O3-2 + 3 H2O --> 4 F- + 2 SO3-2 + 6 H+

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

2 F2 + S2O3-2 + 3 H2O + 6 OH- --> 4 F- + 2 SO3-2 + 6 H+ + 6 OH-

Combine H+ and OH- to form water

2 F2 + S2O3-2 + 3 H2O + 6 OH- --> 4 F- + 2 SO3-2 + 6 H2O

Remove common H2O from both sides

Balanced Eqn is

2 F2 + S2O3-2 + 6 OH- --> 4 F- + 2 SO3-2 + 3 H2O

This is balanced chemical equation in basic medium

Answer:

Coefficients are 2, 1, 4, 2

Water appear as a product with coefficient of 3

4 electrons are transferred

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