## Test 1. Q 2

Posts: 27
Joined: Fri Apr 06, 2018 11:04 am

### Test 1. Q 2

One industrial application of an electrolyte solution is electroplating of metals. In this process, an aqueous solution of metal ions is prepared and electrochemistry can be performed to deposit metal to a desired electrode surface. Suppose one is depositing nickel to an electrode surface and has prepared 50.0 mL of 0.200M of nickel chloride,NiCl2, aqueous solution. After the electrodeposition process, one finds that 0.400g of nickel is now deposited onto the surface. What is the concentration of the nickel ions in the electrolyte now?

So I first solved for mol of Ni, so I did mass of Ni/molar mass of Ni or 0.400g/58.69g/mol, getting 0.007 mol. I then did 0.007 mol/0.05 L to find the concentration but this is incorrect. How were we supposed to solve this problem?

AnnaYan_1l
Posts: 96
Joined: Fri Apr 06, 2018 11:05 am
Been upvoted: 1 time

### Re: Test 1. Q 2

I did not have the same version of the test, but here is what I think are the correct steps:

So, it says that 50.0 mL of .200M of NiCl2 aqueous solution is prepared (called electrolyte). An electrode surface is placed into the solution, and after the electrodeposition process, some of the Ni is deposited on the electrode surface. This means that some Ni from the NiCl2 solution are now plated on the electrode surface (and thus removed from the aqueous solution).

First, find how many moles were in the original aqueous solution: n = Mv = (.200 M NiCl2)*(.050L) = .01 mol NiCl2

Then, determine how many moles of Ni are within that NiCl2: $.01 mol NiCl2 * \frac{1 mol Ni}{1 mol NiCl2} = .01 mol Ni$

Next, determine how many moles of Ni are plated on the surface of the electrode after the electrodeposition process: $.400 g Ni * \frac{1 mol Ni}{58.6934 g Ni} = .00682 mol Ni$

Finally, determine the molarity of the Ni in the electrolyte solution after the electrodeposition process: $\frac{.01 mol Ni - .00682 mol Ni}{.050L} = .0637 M Ni = 6.37 * 10^{-2} M Ni$

I hope this helps! See the image attached with the written out steps that I did, if the text-version is confusing!
Attachments

princessturner1G
Posts: 31
Joined: Fri Apr 06, 2018 11:05 am

### Re: Test 1. Q 2

AnnaYan_1B wrote:I did not have the same version of the test, but here is what I think are the correct steps:

So, it says that 50.0 mL of .200M of NiCl2 aqueous solution is prepared (called electrolyte). An electrode surface is placed into the solution, and after the electrodeposition process, some of the Ni is deposited on the electrode surface. This means that some Ni from the NiCl2 solution are now plated on the electrode surface (and thus removed from the aqueous solution).

First, find how many moles were in the original aqueous solution: n = Mv = (.200 M NiCl2)*(.050L) = .01 mol NiCl2

Then, determine how many moles of Ni are within that NiCl2: $.01 mol NiCl2 * \frac{1 mol Ni}{1 mol NiCl2} = .01 mol Ni$

Next, determine how many moles of Ni are plated on the surface of the electrode after the electrodeposition process: $.400 g Ni * \frac{1 mol Ni}{58.6934 g Ni} = .00682 mol Ni$

Finally, determine the molarity of the Ni in the electrolyte solution after the electrodeposition process: $\frac{.01 mol Ni - .00682 mol Ni}{.050L} = .0637 M Ni = 6.37 * 10^{-2} M Ni$

I hope this helps! See the image attached with the written out steps that I did, if the text-version is confusing!

This was super helpful! How did you know to subtract the molarity instead of using M(initial)V(initial)=M(final)V(final)?

AnnaYan_1l
Posts: 96
Joined: Fri Apr 06, 2018 11:05 am
Been upvoted: 1 time

### Re: Test 1. Q 2

I didn't use M(initial)V(initial)=M(final)V(final) because more solution wasn't being added to the aqueous solution. The idea of M(initial)V(initial)=M(final)V(final) is that you would be adding more solution with NiCl2 (for example), and the original solution would still be in the beaker (if that makes sense). Since Ni was being removed from the beaker and being used to plate the electrode, this is a different situation than would be described using M(initial)V(initial)=M(final)V(final).

Let me know if my explanation didn't make sense, and I can try to explain it differently!