# Find Cell Potential Given Mass

• davev
In summary, the galvanic cell will have a potential of 0.0889 V when the mass of the Pb electrode is 137.08 g.
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?

## Homework Equations

Ecell = Eocell - (RT/nF) * ln(Q)
R = 8.314
n = 2
Q = [products]x/[reactants]y and in this case x and y are both 1
Eocell = 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!

davev said:
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+?

Borek said:
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?

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

Borek said:
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?

davev said:
137.08 g Pb

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

Borek said:
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+.

davev said:
Isn't that what I did?

No, that's not what you did.

Borek said:
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.

Borek said:
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.

Last edited:
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.

davev said:
lnQ = [1.3708/1]

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

## 1. What is cell potential?

Cell potential, also known as cell voltage, is a measure of the potential difference between two electrodes in an electrochemical cell. It is a measure of the energy available to drive an electric current through the cell.

## 2. How is cell potential measured?

Cell potential is measured in volts (V) using a voltmeter. The positive and negative terminals of the voltmeter are connected to the two electrodes of the cell, and the reading on the voltmeter is the cell potential.

## 3. What is the formula for calculating cell potential?

The formula for calculating cell potential is Ecell = Ecathode - Eanode, where Ecathode is the reduction potential of the cathode and Eanode is the reduction potential of the anode. This formula is based on the principles of thermodynamics and the Nernst equation.

## 4. How does mass affect cell potential?

The mass of the electrodes does not directly affect the cell potential. However, the surface area of the electrodes can affect the rate of the electrochemical reactions, which in turn can affect the cell potential. Generally, larger surface area can increase the rate of reaction and therefore increase the cell potential.

## 5. What are some factors that can influence cell potential?

Some factors that can influence cell potential include temperature, concentration of reactants, and the type of electrolyte used in the cell. Changes in these factors can affect the equilibrium between the anode and cathode reactions, resulting in a change in cell potential.

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