Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible

Q.17: Using the standard electrode potentials given in Table 3.1, predict if the reaction between the following is feasible:

(i) Fe3+(aq) and I−(aq)

 (ii) Ag+ (aq) and Cu(s)

(iii) Fe3+ (aq) and Br−(aq)

(iv) Ag(s) and Fe3+(aq)

(v) Br2 (aq) and Fe2+ (aq).

Solution:

For a feasible reaction

∆_rG^θ  < 0

And  ∆_rG^θ= – nFE^o_cell   so that

– nFE^o_cell  <  0

n and F both are always positive values
so that

–E^o_cell  <  0

Change the sign we get

E^o_cell     >  0

Hence for any feasible reaction E^o_cell will always positive

(i)

Use this link to get all values of E^o

http://ncerthelp.blogspot.com/2013/04/the–standard–electrode–potentials–at.html

Balance possible half reactions are

2Fe³⁺ + 2e^–  ––– > 2Fe ⁺²                     E^o= +0.77 V

2I^–                ––– >  I₂ + 2e^–                  E^o   = –0.54 V

Add the values we get

E^o_cell  = +0.23

Here E^o_cell > 0 so reaction is possible

(ii)

2Ag⁺ + 2e^–   ––– > 2Ag                        E^o = +0.80 V

Cu               ––– >  Cu⁺² + 2e^–             E^o   = –0.34 V

Add the values we get

E^o_cell  = +0.46V

Here E^o_cell > 0 so reaction is possible

(iii)

2Fe³⁺ + 2e^–  ––– > 2Fe ⁺²                     E^o = +0.77 V

2Br^–                       ––– >  Br₂ + 2e^–               E^o = –1.09 V

Add the values we get

E^o_cell  = –0.32V

Here E^o_cell <  0 so reaction is not possible

(iv)

2Ag              ––– > 2Ag ⁺¹ + 2e^–            E^o= – 0.80 V

2Fe³⁺ + 2e^–  ––– > 2Fe ⁺²                     E^o = +0.77 V

Add the values we get

E^o_cell  = –0.03V

Here E^o_cell <  0 so reaction is not possible

(V)

Br₂ + 2e^–      ––– > 2Br ^–            E^o = +1.09 V

2Fe²⁺           ––– >  2Fe³⁺ + 2e^– E^o = – 0.77 V

Add the values we get

E^o_cell  = +0.32V

Here E^o_cell > 0 so reaction is possible