Calculate the Standard Cell Potentials of Galvanic Cells in Which Exercise Chapter 3 Electrochemistry Chemistry Class

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 =∆rG^θ and equilibrium constant of the reactions.

Answer:(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°cell = E° right  – E°left
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E°cell = E° Cd  – R°Cr
E°cell = – 0.40 –( – 0.74 )
E°cell = – 0.40 –( – 0.74 )
E°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°cell = + 0.34 V
Use formula
∆rGθ = – nFE°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°cell = E° right  – E°left
Click here to get all values
E°cell = 0.80 – 0.77
E°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°cell = + 0.03 V
Use formula
∆rGθ = – nFE°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