Question

A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 5.20×10⁻²M and 1.50 M, respectively.

A) What is the cell potential when the concentration of Cu2+ has fallen to 0.250 M ?

B) What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.36 V?

Answer 1

Cu²⁺ + 2e^- ------>  Cu, E⁰= +0.34 V
Pb²⁺ + 2e^- -------> Pb, E⁰= - 0.13 V

The overall cell reaction will be as:

Pb ---> Pb²⁺ + 2e^- E⁰= +0.13 V (reversed)
Cu²⁺ + 2e^- ----> Cu, E⁰ = +0.34 V
-------------------------------------------------------------
Pb + Cu²⁺ -----> Pb²⁺ + Cu , E⁰_cell = + 0.47 V

If [Cu²⁺] falls, then [Pb²⁺] increases since it is being produced at the anode.

Now, _______________Pb _____+ _______Cu²⁺ -----> Pb²⁺_____+ _________Cu

Initial_________________________________1.50______0.0520

Change_____________________-(1.50-0.250)=-1.25_____+1.25

Final________________________________0.25_________1.302

Therefore, Q = 1.302 / 0.25 = 5.208 = 5.21

Now, E_cell = E⁰_cell - 0.059/n *logQ

Here n= 2 (since two electrons are transferred)

=> E_cell = 0.47 - 0.059/2 *log 5.21

=> E_cell = 0.47 -0.0295 x 0.7168

=>  E_cell = 0.47 -0.0211456

=> E_cell = 0.448 V = 0.45 V

A) Hence, the  cell potential when the concentration of Cu2+ has fallen to 0.250 M is  0.45 V.

B)

We must note that sum of [Pb²⁺] and [Cu²⁺] = 0.0520 + 1.50 = 1.552

Again,

E_cell = E⁰_cell - 0.059/n *logQ

=> 0.36 = 0.47 - 0.059/2*logQ

=> 0.36-0.47= -0.0295 *log Q

=> -0.11 = -0.0295 *log Q

=> log Q = 0.11/0.0295

=> log Q = 3.73

=> Q = antilog (3.73)

=> Q = 5370

If x = [Pb²⁺] and y = [Cu²⁺], then x + y = 1.552 and x/y = 5370 => x = 5370 y

Now,

x + y = 1.552

=> 5370 y + y = 1.552

=> 5371 y = 1.552

=> y = 1.552/5371

=> y = 0.0002889 = [Cu²⁺]

And x = [Pb²⁺] = 5370y = 5370 x 0.0002889 = 1.55

Therefore,  [Cu²⁺] = 0.0002889 and [Pb²⁺] = 1.55