Using the Standard Electrode Potentials Given in Table 3.1, Predi Exercise Chapter 3 Electrochemistry Chemistry Class

Question 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).

Answer:For a feasible reaction
∆rG^θ  < 0
And  ∆rG^θ = – nFE°cell   so that
– nFE°cell  <  0
n and F both are always positive values
so that
–E°cell    <  0
Change the sign we get
E°cell      >  0
Hence for any feasible reaction E°cell will always positive
(i)
Use this link to get all values of E°
http://ncerthelp.blogspot.in/2013/04/the-standard-electrode-potentials-at.html
Balance possible half reactions are
2Fe³⁺ + 2e^-  →2Fe⁺²                 E° = +0.77 V
2I^–                → I₂ + 2e^-              E°   = –0.54 V
Add the values we get
E°cell  = +0.23
Here E°cell > 0 so reaction is possible
(ii)
2Ag⁺ + 2e^-   →2Ag                              E° = +0.80 V
Cu               → Cu⁺² + 2e^-                   E°   = –0.34 V
Add the values we get
E°cell  = +0.46V
Here E°cell > 0 so reaction is possible
(iii)
2Fe³⁺ + 2e^-  →2Fe⁺²                 E° = +0.77 V
2Br^–                       → Br₂ + 2e^-           E° = –1.09 V
Add the values we get
E°cell  = –0.32V
Here E°cell <  0 so reaction is not possible
(iv)
2Ag              →2Ag⁺¹ + 2e^-                  E° = – 0.80 V
2Fe³⁺ + 2e^-  →2Fe⁺²                 E° = +0.77 V
Add the values we get
E°cell  = –0.03V
Here E°cell <  0 so reaction is not possible
(V)
Br₂ + 2e^-      →2Br^–                  E° = +1.09 V
2Fe²⁺           → 2Fe³⁺ + 2e^-        E° = – 0.77 V
Add the values we get
E°cell  = +0.32V
Here E°cell > 0 so reaction is possible