Question

In: Chemistry

Given the following pairs of half-reactions, please construct a complete balanced oxidation-reduction reaction for each pair...

Given the following pairs of half-reactions, please construct a complete balanced oxidation-reduction reaction for each pair of two half-reactions:

a) H2= 2H+ + 2e-

Fe2+= à Fe3+ + e-

b) 1/4O2(g) + H++ e- == ½ H2OH2 =à 2H++ 2e-

   C) 2IO3-+ 12H++ 10e-== I2+ 6H2O 2I-= I2+ 2e-

Solutions

Expert Solution

the value of comparison i in alll three questions taken from standard reduction potential table

1) H2------> 2H+ + 2e- -----1

  Fe2+-------> à Fe3+ + e- -------2

multiply equation 2 by 2

2 Fe2+-------> 2 Fe3+ + 2e-

the reaction are in their oxidation state changing them to reduction

2H+ + 2e- ------> H2 ,   2 Fe3+ + 2e- -------> 2 Fe2+

as the reductionn potential of   2 Fe3+ + 2e- -------> 2 Fe2+ reaction is having a positive value that means it act as

cathode (+) while the H2 ------> 2H+ + 2e -     act as anode (-)

on solvong these 2 equation we get

2 Fe3+ + H2 --------> 2H+ + Fe2+

but the reaction of    Fe2+-------> à Fe3+ + e- is given in reverse order it means that the E0reduction value for reverse reaction will become -ve it means its reduction potenstial value is more negative then H2 reaction now

2H+ + 2e- -----> H2 act as cathode (+)

2Fe2+-------> 2Fe3+ + e- act as anode (-)

Fe2+ +   2H+   ------>  H2 + 2Fe3+

2)  1/4O2 (g) + H+ + e- ------> 1/2 H2O eq 1 , H2 ------> à 2H+ + 2e- eq2

as the reductionn potential of 1/4O2 (g) + H+ + e- ------> 1/2 H2O is greater it means it act as cathode (+)

while the H2 ------> à 2H+ + 2e act as anode (-)

mutilply eq 1 by 2 we get

1/2 O2 (g) + 2H+ + 2e- ------> H2O

H2 ------> à 2H+ + 2e-

on solving we get

1/2 O2 + H2 ----->  H2O

3)  2IO3- + 12H+ + 10e- -----> I2 + 6H2O ,

2I- -------->I2 + 2e-

multiply eqn 2 by 5

10 I - --------> 5 I2 + 10 e-     

as the reduction potential for the 2IO3- + 12H+ + 10e- -----> I2 + 6H2O is more positive that means it is

cathode +ve value and for the reaction   10 I - --------> 5 I2 + 10 e- it is anode (-ve)

we get

2IO3- + 12H+ + 10 I - ----------> 5 I2 + I2 + 6H2O

the value of comparison is taken from standard reduction potential table


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