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Question 4

Calculate the standard cell potentials of galvanic cells in which the following reactions take place:

(i) 2Cr(s) + 3Cd2+(aq) → 2Cr3+(aq) + 3Cd

(ii) Fe2+(aq) + Ag+(aq) → Fe3+(aq) + Ag(s)

Calculate the ΔrGø¸ and equilibrium constant of the reactions.

Answer

(i) Eø Cr3+ / Cr  = - 0.74 V

   Eø Cd2+ / Cd  = - 0.40 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

Eø  = EøR  - EøL

     = -0.40 - (-0.74)

     = + 0.34 V

ΔrGø = -nFEøcell

In the given equation,

n = 6

F = 96487 C mol - 1

Eøcell = +0.34 V

Then, ΔrGø  = - 6 × 96487 C mol - 1 × 0.34 V

= - 196833.48 CV mol - 1

= - 196833.48 J mol - 1

= - 196.83 kJ mol - 1

Again,

ΔrGø = - RT ln K

= 34.496

K = antilog (34.496)

= 3.13 × 1034

 

(ii) Eø Fe3+ / Fe2+  =  0.77 V

   Eø Ag+ / Ag        =  0.80 V

The galvanic cell of the given reaction is depicted as:

Now, the standard cell potential is

Eø  = EøR  - EøL

       = 0.80 - 0.77

     = 0.03 L

Here, n = 1.

Then, ΔrGø = -nFEøcell

= - 1 × 96487 C mol - 1 × 0.03 V

= - 2894.61 J mol - 1

= - 2.89 kJ mol - 1

Again, ΔrGø = - 2.303 RT ln K

= 0.5073

K = antilog (0.5073)

= 3.2 (approximately)

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