6N. 7a

[Lilyana Villa 3L]

6N.7) "Calculate Ecell for each of the following concentration cells: a) Cu(s)|Cu 2+ (aq, 0.0010 mol ⋅ L^−1 )||Cu 2+ (aq, 0.010 mol ⋅ L^−1 )|Cu(s) ." Could someone explain how you would do this problem? I don't understand how you would create the redox reaction from this cell diagram since both the anode and cathode are the same reaction. Wouldn't Cu2+ and Cu just cancel out? I hope this question made sense.

[Manseej Khatri 2B]

Hi. Even though the half reactions are the same, the concentrations are different.
You can calculate the Ecell for non standard conditions by E(cell) = E(naught) -(RT/nF)lnQ where Q is the equilibrium quotient.
Note the cathode concentration would be considered the reactant concentration, and the anode concentration would be considered the product concentration.

[Lilyana Villa 3L]

Hi. Even though the half reactions are the same, the concentrations are different.
You can calculate the Ecell for non standard conditions by E(cell) = E(naught) -(RT/nF)lnQ where Q is the equilibrium quotient.
Note the cathode concentration would be considered the reactant concentration, and the anode concentration would be considered the product concentration.

Would the cathode concentration always be considered the reactant concentration and the anode concentration always the product concentration?

[chinmayeec 2H]

Hi. Even though the half reactions are the same, the concentrations are different.
You can calculate the Ecell for non standard conditions by E(cell) = E(naught) -(RT/nF)lnQ where Q is the equilibrium quotient.
Note the cathode concentration would be considered the reactant concentration, and the anode concentration would be considered the product concentration.

Would the cathode concentration always be considered the reactant concentration and the anode concentration always the product concentration?

No, not necessarily. For example, in this reaction:

2Cl- -> Cl2 +2e-
2H+ +2e- ->H2
Overall Reaction: 2Cl- +2H+ --> H2 + Cl2

K=[H2][Cl2]/[Cl-]^2[H+]^2
Here, you can see that Cl- is part of the anode but it is one of the reactants in the overall equation. I suggest always writing out the overall reaction to determine the equilibrium expression.

[Lilyana Villa 3L]

Hi. Even though the half reactions are the same, the concentrations are different.
You can calculate the Ecell for non standard conditions by E(cell) = E(naught) -(RT/nF)lnQ where Q is the equilibrium quotient.
Note the cathode concentration would be considered the reactant concentration, and the anode concentration would be considered the product concentration.

Would the cathode concentration always be considered the reactant concentration and the anode concentration always the product concentration?

No, not necessarily. For example, in this reaction:

2Cl- -> Cl2 +2e-
2H+ +2e- ->H2
Overall Reaction: 2Cl- +2H+ --> H2 + Cl2

K=[H2][Cl2]/[Cl-]^2[H+]^2
Here, you can see that Cl- is part of the anode but it is one of the reactants in the overall equation. I suggest always writing out the overall reaction to determine the equilibrium expression.

Okay that makes sense! So would what I said in my previous statement only pertain to concentration cells? Or just this specific problem?