In: Chemistry
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-
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