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davev

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## 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?Pb2+ + 2 e- → Pb Eo = -0.133 V

__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!