Q. Write a balanced redox equation,and calculate E o for; A) The reaction betwee

Question

Q. Write a balanced redox equation,and calculate Eo for;

A) The reaction between VO2+(aq) and Na2SO3 in acid solution. (Comment on the spontaneity or otherwise of this reaction and the need to heat the reaction mixture.)

B) The reaction between VO2+(aq) and NaCl in acid solution (Comment on the spontaneity or otherwise of this reaction)

Given.

VO2+ + 4H+ + 3e => V2+ + 2H2O        (1.1V)

Zn2+ + 2e => Zn                               (-0.76V)

O2 + 4H+ + 4e => 2H2O                     (1.23V)

V3+ + e => V2+                                   (-0.26V)

VO2+ + 2H+   + e => VO2+ + H2O     (1.00V)

SO42- + 4H+ + 2e => H2SO3 + H2O (0.17V)

Cl2 + 2e => 2Cl-                                   (1.35V)

Explanation / Answer

1. VO2+(aq) and Na2SO3 in acid solution

Cathode(Reduction):VO2+ + 4H+ + 3e- ---> V2+ + 2H2O        (Ecell0 = 1.1V)

Anode (Oxidation): SO32– + H2O --->SO42– + 2H+ + 2e- (Ecell0 =0.17 V)

Ecell0 = Ecathode0- Eanode0 = 1.1 V - 0.17 V = 0.93 V

Ecell0 is positive hence the reaction would be spontaneous.

Reduction:(VO2+ + 4H+ + 3e- ---> V2+ + 2H2O ) * 2 => 2VO2+ + 8H+ + 6e- ---> 2V2+ + 4H2O

Oxidation: (SO32– + H2O --->SO42– + 2H+ + 2e- ) *3 => 3SO32– + 3H2O --->3SO42– + 6H+ +6e-

Adding half reactions,

2VO2+ + 8H+ + 6e- ---> 2V2+ + 4H2O

3SO32– + 3H2O --->3SO42– + 6H+ +6e-

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2VO2+ + 3 SO32– + 2H+ ---->2V2+ + 3SO42– + H2O

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VO2+(aq) and NaCl in acid solution

Cathode(Reduction):VO2+ + 4H+ + 3e- ---> V2+ + 2H2O        (Ecell0 = 1.1V)

Anode (Oxidation): 2Cl- ---> Cl2 + 2e- (Ecell0 = 1.35V)

Ecell0 = Ecathode0- Eanode0 = 1.1 V - 1.35 V = -0.25 V

Ecell0 is negative hence the reaction would be non spontaneous.

Reduction:(VO2+ + 4H+ + 3e- ---> V2+ + 2H2O ) * 2 =>   2VO2+ + 8H+ + 6e- ---> 2V2+ + 4H2O

Oxidation: (2Cl- --->  Cl2 + 2e-  ) *3 => 6Cl- ---> 3Cl2 + 6e-  

Adding half reactions,

2VO2+ + 8H+ + 6e- ---> 2V2+ + 4H2O

6Cl- ---> 3Cl2 + 6e-

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2VO2+ + 8H+ + 6Cl----->2V2+ + 4H2O +3Cl2

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