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

Q.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 =∆_rG^θ and equilibrium constant of the reactions.

Solution:

(i)                  

The galvanic cell of the given reaction is represented as

Cr(s) | Cr³⁺ || Cd²⁺ | Cd(s)

The formula of standard cell potential is

E^o_cell = E^o _right  – E^o_left

Use this link to get all values

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

E^o_cell = E^o _Cd  – R^o_Cr

E^o_cell = – 0.40 –( – 0.74 )

E^o_cell = – 0.40 –( – 0.74 )

E^o_cell =  + 0.43 V

In balanced reaction there are 6 electron are transferring so that n = 6

Faraday constant, F = 96500 C mol⁻¹

E^o_cell = + 0.34 V

Use formula

∆_rG^θ = – nFE^o_cell

Plug the value we get   

Then, = −6 × 96500 C mol⁻¹ × 0.34 V

= −196860 CV mol−1

= −196860J mol−1

= −196.86 kJ mol−1

Again,

Use second formula of ∆_rG^θ

∆_rG^θ = −2.303RT log k_C

log K_C = (∆_rG^θ) /( – 2.303RT)

plug the values we get

 (ii)

The galvanic cell of the given reaction is represented as

Fe²⁺(aq) | Fe³⁺(aq) || Ag⁺ | Ag(s)

The formula of standard cell potential is

E^o_cell = E^o _right  – E^o_left

Use this link to get all values

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

E^o_cell = 0.80 – 0.77

E^o_cell =  + 0.03 V

In balanced reaction there are 1 electron are transferring so that n = 1

Faraday constant, F = 96500 C mol⁻¹

E^o_cell = + 0.03 V

Use formula

∆_rG^θ = – nFE^o_cell

Plug the value we get   

Then, = −1 × 96500 C mol⁻¹ × 0.03 V

= −2895 CV mol⁻¹

= −2895J mol⁻¹

= −2.895 kJ mol⁻¹

Again,

Use second formula of ∆_rG^θ

∆_rG^θ = −2.303RT log k_C

log K_C = (∆_rG^θ) /( – 2.303RT)

plug the values we get