## 6N3.B

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Alex Tchekanov Dis 2k
Posts: 118
Joined: Sat Aug 24, 2019 12:16 am

### 6N3.B

Could someone please explain how we know which element is the product or reactant in this problem? I did everything correctly but instead, I put Zinc as the reactant when I solved for Q and I got the wrong answer. I assumed Zinc was the reactant because it is the anode in this equation and that's where the electrons will come from...

Benjamin Feng 1B
Posts: 102
Joined: Sat Sep 07, 2019 12:19 am

### Re: 6N3.B

You are right that zinc is the reactant, but that means that Zn2+ is the product, so that goes on top when calculating Q

Julie_Reyes1B
Posts: 105
Joined: Sat Jul 20, 2019 12:16 am

### Re: 6N3.B

For that problem, it does look like Zinc is a reactant and the zinc ion is the product. It seems like you got that part right! Maybe you made a mistake with the Q term; for that problem, make sure it is [Zn2+]/[Ni2+].

In terms of deciding which is the anode and cathode, the locations of the substances in the cell diagram will tell you. The right tends to be the cathode, and the left tends to be in the anode. If you're unsure, you can also look at the potentials and the one with a stronger reduction potential (will be reduced) will be the cathode.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

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