posted by Cori on .
a glavanic cell is set up by connecting the following half-cells together: nickle electrode in nickel nitrate solution and a copper electrode in copper nitrate solution.
in which half-cell will oxidation occur?
in which half-cell will reduction occur?
write the half reaction for both
can someone please help me I have been trying at this question for almost an hour and can't figure it out please help me, thank you.
Look up the value of Eo for Ni and Cu.
You want Ni^+2 + 2e ==> Ni Eo = ??
and Cu^+2 + 2e ==> Cu Eo = ??
Look at the values, choose the one with the LARGER negative value, reverse that equation and change the sign from - to + for the Eo value, then add the two equations to get the cell reaction. Add the two Eo values to obtain the cell potential. Then write the reaction at each electrode, and note that oxidation occurs at the. That electrode will be the negative electrode. Post your work if you get stuck.
I'm sorry, but how do I find Eo?
Look in your text or notes. You should find a table called "Standard Reduction Potentials" or something like that. In most texts it is a rather lengthy 3 or 4 pages and you will need to look at each reaction to find the two you want. I can help a little. Look in the -0.34 range for the copper couple and look in the 0.-25 for the nickel couple.
I found the EO, I am a little stuck now, you said to choose the one with the larger negative value, but they are both posisitve so would I choose the smaller one?
This is what I found:
Ni2+(aq) + 2 e- -----> Ni(s)
Cu2+(aq) + e- -----> Cu+(aq)
I can't see your table so I don't know what you've found; however, written as a REDUCTION (Ni2+ + 2e==> Ni(s) the value of Eo will be -0.250 or close to that (not positive). Make sure the table you are using writes the reaction as a reduction. If the equation is written as an oxidation (Ni ==> Ni2+ + 2e), then the value will be 0.250 or close to that value. If it is written as an oxidation, then turn it around to get the - number as a reduction. The corresponding value for Cu is -0.-337 (again, not positive) for the reduction potential.
Well, I am wrong. I looked one up wrong. Here is a table.
The Cu^+2 + 2e ==> Cu Eo = 0.34
Ni^+2 + 2e ==> Ni(s) Eo = -0.250
Use these values as is without reversing anything. Follow the other instructions in my previous post.
Okay thank you.
sorry to bother you again, but I understand how to add the values, but how do you add the equations, i understand everything else except for adding the equations, as i am not sure how to do that. Thanks.
Post your answers if you want me to check them. I'm sorry I looked the values up wrong the first time.
Let me start over to make sure we are on the same page.
Here is the reduction equation for Cu and the potential.
Cu^+2 + 2e ==> Cu(s) Eo = +0.337 (notice that what you looked up was Cu+2 going to Cu+1.
The reduction potential for Ni and the equation is
Ni^+2 + 2e ==> Ni(s) Eo = -.250
Obviously, we can't add two equations written as all reductions or all oxidations. The reaction is between an oxidation half cell and a reduction half cell. So we take the more negative value (Ni) and reverse it so the two equations now read as follows:
Cu^+2 + 2e ==> Cu(s) E(red) = 0.337
Ni(s) ==> Ni^+2 + 2e E(ox) = +0.250
Then we add the equations and the E values.
The equation is
Cu^+2 + 2e + Ni(s) ==>Cu(s) + Ni^+2 + 2e
Cancel the 2e common to both sides and you have the final equation.
The voltage is the sum of the two half equations (one ox and the other red) so the voltage of the cell is 0.337 + 0.250 = ??
I hope I haven't totally confused you. To be honest about it., I looked the values up in a table of OXIDATION potentials and forgot to reverse one of them.
So the electrode reactions are,
At the Ni, it is
Ni==> Ni^+2 + 2e and it is negative because it is giving up electrons. This is the anode, by definition, for this is where oxidation is occurring.
At the Cu electrode, the reaction is
Cu^+2 + 2e ==> Cu and that electrode is positive because it is accepting electrons.
Let me know if you have any questions. After a short while, I shall delete those previous posts where I gave conflicting information. I don't want to leave bad info on the board forever. I hope this helps.
Thank you so much for your help. I appreciate it so much. Once again thank you, I am so greatful. Thanks!