Question

1. What is the calculated value of the cell potential at 298K for an electrochemical cell with the following reaction, when the Cl₂ pressure is 9.21×10^-4 atm, the Cl^- concentration is 1.38M, and the Pb²⁺ concentration is 1.19M ?

Cl₂(g) + Pb(s) -------> 2Cl^-(aq) + Pb²⁺(aq)

Answer: _______ V

The cell reaction as written above is spontaneous for the concentrations True or False?

(This is all part of one question)

Answer 1

Reaction at anode:Pb _(s)   Pb ²⁺ (aq) + 2 e ^-

Reaction at cathode: Cl _{2 (g)} + 2 e ^-   2 CI ^-_(aq)

Here n= no of electrons transferred = 2

Q = [ Pb ²⁺ ] [ 2 CI ^- ] / ( P Cl ₂ )

E ⁰_c = E ⁰ cathode - E ⁰ anode

= 1.36 V - ( - 0.126 V)

= 1.486 V

Emf of above cell is calculated by using Nernst equation as

E cell = E ⁰ cell - [ 2.303 R T / n F ] log Q

Where , n is no of electrons transferred in the reaction ,

Q is the reaction quotient.

E cell = 1.486 - 2.303 R T / n F log Q

= 1.486 - 0.0591 / 2 log [ Pb ²⁺ ] [ CI ^- ] ² / ( P Cl ₂ )

= 1.486 - 0.02955 log 1.19  ( 1.38 ) ² / ( 9.21 x 10 ^-04 )

= 1.486 - 0.02955 x 3.3910

= 1.486 - 0.100

= 1.386 V

Reaction is spontaneous .

ANSWER : TRUE

E cell > 0 and hence G ⁰ < 0 . So reaction is spontaneous.