What Is the Cell Potential When Pb Electrode Mass Changes?

[davev]

Homework Statement

Consider the following half reactions at 298 K

Ni2+ + 2 e- → Ni Eo = -0.231 V
Pb2+ + 2 e- → Pb Eo = -0.133 V​

A galvanic cell based on these half reactions is set up under standard conditions where each solution is 1.00 L and each electrode weighs exactly 100.0 g. What will the potential of the cell be when the mass of the Pb electrode is 137.08 g?

Answer: 0.0934 V

Homework Equations

E_cell = E^o_cell - (RT/nF) * ln(Q)
R = 8.314
n = 2
Q = [products]^x/[reactants]^y and in this case x and y are both 1
E^o_cell = 0.098 V

Pb2+ + Ni → Pb + Ni2+
Pb2+ is the cathode (reduction); Ni2+ is the anode (oxidation).

The Attempt at a Solution

I need some guidance on how to deal with the mass in grams.

Originally, as I was given the volume, I was going to use molar mass to get moles of Ni and PB, then divide by the volume given to get the concentration of each. Instead of using the 100 grams of Pb, I used 137.08 grams to find the concentration -- keeping the concentration found using 100 grams of Ni as is. This was wrong when I plugged my numbers into the Nernst equation.

My second thought was to set up an ICE table; however, the original concentration using 100 grams at the cathode (if using the molar mass to find moles and using volume to find concentration is correct) is less than the new concentration using 137.08 grams, so I don't see how an ICE table would work, because the cathode concentration should always decrease.

Any help would be great!

 

[Borek]

A galvanic cell based on these half reactions is set up under standard conditions

What does the "standard conditions" tell you about initial concentrations of Ni²⁺ and Pb²⁺?

 

[davev]

What does the "standard conditions" tell you about initial concentrations of Ni²⁺ and Pb²⁺?

That the concentration of both anode and cathode is 1 M. How would I incorporate the change in mass of the cathode electrode though?

 

[Borek]

Simple stoichiometry - moles, masses, reaction equations, these things.

 

[davev]

Simple stoichiometry - moles, masses, reaction equations, these things.

I'm confused, because I don't think I get the right answer by doing this:

137.08 g Pb * (1 mol Pb/207.2 g Pb) * (1 mol Pb2+/1 mol Pb) = 0.66158 moles of Pb2+. I'm assuming that is also 0.66158 M of Pb2+ as there is 1 L of solution for both anode and cathode.

1 - x = 0.66158
x = 0.33842; this is how much the concentration of both would change.

Then lnQ = (1.33842/0.66158). Multiply this by (8.314*298)/(2*96485). Finally 0.098 - 0.009046564 = 0.0889 V, which is not the correct answer. What am I doing wrong?

 

[Borek]

137.08 g Pb

That's the final mass, not the mass deposited.

 

[davev]

That's the final mass, not the mass deposited.

Isn't that what I did? I used that as the end product in my ICE table. To find change (i.e., x), 1 - x = 0.66158. The value of x would be the final concentration of Ni2+.

 

[Borek]

Isn't that what I did?

No, that's not what you did.

 

[davev]

No, that's not what you did.

Okay — could you explain what I have to do more clearly, because I don't have any other ideas on how to approach this problem.

 

[davev]

No, that's not what you did.

I figured it out, but I'm somewhat confused. lnQ = [1.3708/1] is the right way to approach this problem. However, why doesn't the cathode concentration change at all?

Edit: Sorry, this is wrong. It's closer than I've come before to the actual answer, but it's wrong. I'm assuming the cathode definitely does have to change.

 

[Borek]

You are interested only in the mass of lead that was deposited, not in the lead that was there (in the electrode) from the very beginning. Whatever was deposited, came from the solution. Whatever was there - just was there, and its presence doesn't count against changes in the solution composition.

lnQ = [1.3708/1]

And how on Earth do you expect me to guess what these random numbers means?