Question

This question has already been answered, but there was not as much detail as i needed to understand what exactly to do and i need to understand it quick... could you please walk me through how to do this, but specifically A and B please plug in all numbers and state where they came from , thank you so much!

Calculate the potential in the following cell

Pb|Pb2+(1.00M)|| Ag+(1.00M) |Ag

a. right at the beginning when you start using this voltaic cell,

b. when 50% of the Ag+ ions have been consumed,

c. when 99% of the Ag+ ions have been consumed, and

d. when 99.99% of the Ag+ ions have been consumed. Keep in mind that both, the Pb2+ and the Ag+ concentrations are changing.

Answer 1

To calculate the potential of a cell we will first calculate the standard electrode potential of cell

E⁰cell = E⁰cathode - E⁰anode

Here the cathode is Silver and anode is lead

Reduction of potential

Silver = 0.799   Lead = -0.126

E⁰cell = 0.799 - (- 0.126) = 0.925 V

The cell reaction is

2Ag+ + Pb(s) --> Pb+2 + 2Ag(s)

a) right at the beginning when you start using this voltaic cell

Ecell = E⁰cell - 0.0592 / n )log [Pb+2) / (Ag+)^2 ] = 0.925 - 0.0296 log (1/1) = 0.925V

b) when 50% of the Ag+ ions have been consumed,25% of Pb+2 will be produced

so final conc of Ag+ = 0.5 M

And that of Pb+2 = 1.25 M

Ecell = E⁰cell - 0.0592 / n )log [Pb+2) / (Ag+)^2 ] = 0.925 - 0.0296 log (1.25/ [0.5]^2) = 0.925 - 0.0296 (-0.698)

= 0.945 V

c) when 99% of the Ag+ ions have been consumed

so we have 1-0.99 of Ag+ = 0.01 M

And we will have 1+0.99 / 2 = 1.495 M

Ecell = E⁰cell - (0.0592 / n )log [Pb+2) / (Ag+)^2 ] = 0.925 - 0.0296 log (1.495/ [0.01]^2) = 0.925 - 0.0296 (4.17)

= 0.925 - 0.123 = 0.802 V

d) when 99.99% of the Ag+ ions have been consumed

We will left with 1-0.999 of Ag+ = 0.001

Pb+2 produced = 1+0.999/2 = 1.499

Ecell = E⁰cell - (0.0592 / n )log [Pb+2) / (Ag+)^2 ] = 0.925 - 0.0296 log (1.499/ [0.001]^2) = 0.925 - 0.0296 (6.175)

= 0.925 - 0.182 = 0.742V