# Electrochemistry

In lab, we did an experiment with electrochemical cells with solutions of ZnSO4, CuSO4, Al2(SO4)3 and MgSO4 and their respective metal electrodes.

There are a couple of post-lab questions that I'm not sure about:
(1) Summarize Part A with a data table containing the headings: Anode, Cathode, Overall Cell Reaction, Cell Potential. For each of cell combination, write the anode half reaction, the cathode half reaction, the overall reaction in cell notation (all concentrations of solutions are 0.1M) and the potential in volts.

Cell Potential is the same as Voltage Potential, right?

I'm not completely sure about writing half reactions, so I wanted to post a couple of the ones I wrote and if anybody could check these over for me, that would be completely helpful:

Cu/Zn
anode: Zn(s) --> Zn^2+(aq) + 2 e-
cathode: Cu^2+(aq) + e- --> Cu(s)

Cu/Mg
anode: Mg(s)---> Mg^2+(aq) + 2 e-
cathode: Cu^2+(aq) + e- --> Cu(s)

Zn/Al
anode: Al(s) --> Al^3+(aq) + 3 e-
cathode: Zn^2+(aq) + 2 e- --> Zn(s)

Cu/Al
For example, for the cell combination of Cu/Zn I got a voltage potential of 0.515V. So is this the cell potential?

I'm not really clear on how to write overall reactions either. Would these be correct?

Cu/Zn
Zn(s)|Zn^2+(aq, 0.1M)||Cu^2+(aq, 0.1M)|Cu(s)

Zn/Al
Al(s)|Al^3+(aq, 0.1M)||Zn^2+(aq, 0.1M)|Zn(s)

(2) Create a Reduction Potential Table with the cells containing copper. Assume the reduction potential for Cu^2+ + 2e- --> Cu is 0.00V and calculate the remaining reduction potentials relative to Cu^2+/Cu. List the reactions in order of decreasing reduction potential.

So...I would just compare voltages?
Cu/Zn is 0.515 V
Cu/Mg is 1.728 V
Cu/Al is 0.729 V

Does that mean I make the assumptions that Eo(V)
Zn|Zn^2+ is 0.515 V
Mg|Mg^2+ is 1.728 ?

1. 👍 0
2. 👎 0
3. 👁 1,009
1. n lab, we did an experiment with electrochemical cells with solutions of ZnSO4, CuSO4, Al2(SO4)3 and MgSO4 and their respective metal electrodes.

There are a couple of post-lab questions that I'm not sure about:
(1) Summarize Part A with a data table containing the headings: Anode, Cathode, Overall Cell Reaction, Cell Potential. For each of cell combination, write the anode half reaction, the cathode half reaction, the overall reaction in cell notation (all concentrations of solutions are 0.1M) and the potential in volts.

Cell Potential is the same as Voltage Potential, right?

That is right.

I'm not completely sure about writing half reactions, so I wanted to post a couple of the ones I wrote and if anybody could check these over for me, that would be completely helpful:

Cu/Zn
anode: Zn(s) --> Zn^2+(aq) + 2 e-
cathode: Cu^2+(aq) + e- --> Cu(s)

Cu/Mg
anode: Mg(s)---> Mg^2+(aq) + 2 e-
cathode: Cu^2+(aq) + e- --> Cu(s)

Zn/Al
anode: Al(s) --> Al^3+(aq) + 3 e-
cathode: Zn^2+(aq) + 2 e- --> Zn(s)

All of the half cell reactions are correct and the correct notation (anode or cathode) is given. Good work.

Cu/Al
For example, for the cell combination of Cu/Zn I got a voltage potential of 0.515V. So is this the cell potential?

You didn't show your work here so I can't check it for cell potential.

I'm not really clear on how to write overall reactions either. Would these be correct?

Cu/Zn
Zn(s)|Zn^2+(aq, 0.1M)||Cu^2+(aq, 0.1M)|Cu(s)

Zn/Al
Al(s)|Al^3+(aq, 0.1M)||Zn^2+(aq, 0.1M)|Zn(s)

No, the overall cell reactions are different. What you have done is draw the cell notation showing the anode, cathode, concns., etc. What you want for the overall cell reaction is to take the two half cells and add them together; for example, for the Zn/Cu cell, you had above the following:
Zn ==> Zn^+2 + 2e
Cu^+2 + 2e ==> Cu
Now just add those together to obtain
Zn + Cu^+2 ==> Cu + Zn^+2
Note that I have omitted (aq) and (s).

(2) Create a Reduction Potential Table with the cells containing copper. Assume the reduction potential for Cu^2+ + 2e- --> Cu is 0.00V and calculate the remaining reduction potentials relative to Cu^2+/Cu. List the reactions in order of decreasing reduction potential.

So...I would just compare voltages?
Cu/Zn is 0.515 V
Cu/Mg is 1.728 V
Cu/Al is 0.729 V

No, what they want here, I think, is for you to set Cu equal to zero, and using your measured voltage, calculate the half reactions for each metal RELATIVE TO Cu at ZERO.

Does that mean I make the assumptions that Eo(V)
Zn|Zn^2+ is 0.515 V
Mg|Mg^2+ is 1.728 ?

If you calculated the cell potential of the Zn/Cu cell as 0.515, then the Zn to Zn^+2 + 2e can't be 0. The 0.515 you measured/calculated is relative to hydrogen set to zero. Check my thinking.

1. 👍 0
2. 👎 0
2. Note my very last comment. It should read, If you calculated the cell potential of the Zn/Cu cell as 0.515, then the Zn to Zn^+2 can't be 0.515.The 0.515 you measured.........etc.
If you calculated the cell potential of the Zn/Cu cell as 0.515, then the Zn to Zn^+2 + 2e can't be 0. The 0.515 you measured/calculated is relative to hydrogen set to zero. Check my thinking.

1. 👍 0
2. 👎 0

Cu/Al
For example, for the cell combination of Cu/Zn I got a voltage potential of 0.515V. So is this the cell potential?

You didn't show your work here so I can't check it for cell potential.

^ I didn't do any work, that was the reading that the voltage probe gave.

Should I be using the Nernst equation?

No, the overall cell reactions are different. What you have done is draw the cell notation showing the anode, cathode, concns., etc. What you want for the overall cell reaction is to take the two half cells and add them together;

^ I see what you did, thanks. That made a lot of sense. But the question asks for the overall reaction in cell notation... Is that different from Zn + Cu^+2 ==> Cu + Zn^+2 ?

If you calculated the cell potential of the Zn/Cu cell as 0.515, then the Zn to Zn^+2 + 2e can't be 0. The 0.515 you measured/calculated is relative to hydrogen set to zero. Check my thinking.

^ Now I'm thinking that this is a Nernst equation thing as well.
So usually Eo = 0.00V for 2 H^+(aq) + 2 e- --> H2(g)
but the question is saying that

Eo = 0.00V for Cu^2+ + 2e-

Nernst equation:
Ecell = Eo - (0.059/n) log Q
where n = mol of e-
and Q = reaction quotient

so for Zn/Cu, would it be

Ecell = 0.00 - (0.059 / 2) log Q
I know what reaction quotient is but is that Q of the half reaction Zn --> Zn^2+ + 2e-?
Is Q just 0.1?
Or do I have to take 0.1 and multiply by 2 for the 2e-?

1. 👍 0
2. 👎 0
posted by Donna
4. OK. Sorry to have taken so long but I've not had much time today. Here is the best I can do with your follow up questions.

Concerning the cell potential versus the voltage, now that we have had these conversations, PERHAPS the cell potential was meant to be one column filled in with calculated values using the Nernst equation while the voltage column was to be the measured potential with your voltage probe. I have done a bunch of these BUT I never had a voltage probe. I always used a Wheatstone bridge to measure the voltage because the voltmeters of my day took too much current to operate them; therefore, they always gave voltages that were too low (because the voltmeters took more current to operate them than the cells could deliver and there was a voltage drop). Perhaps new voltage probes are better.

Regarding the cell reaction, I went back and reread the instructions you showed in the first post and it DOES say to write it in cell notation; therefore, what you wrote is correct. Frankly, I have never seen a cell reaction written as a cell notation. I have always written cell reactions as cell reactions and cell notations as cell notations but since your instructions were to use cell notation to show the cell reaction, those look good to me.

Regarding the Cu/Al and similar cells and the cell potential. I can't say if those are the cell potential. I know if we go through the Nernst equation that you don't get anywhere near the values you measured with the probe. Whether it is a faulty probe, a probe that draws too much current, or something again, I don't know because I wasn't there to see what was going on. It may be that you were to calculate the cell potential from the Nernst equation. For example, the Zn/Cu reaction gives a calculated value of 1.099 v and the probe gave 0.515 v.

Regarding the cells with Cu and using Cu^+2 + 2e ==> Cu E = 0.00, here are the listings.
Zn/Cu cell 0.515 v.
Mg/Cu cell 1.728 v.
Al/Cu cell 0.729 v.

I'll do just one of these.
Zn ==> Zn^+2 + 2e ??
Cu^+2 + 2e ==> Cu Eo = 0
===========================
Zn + Cu^+2 ==> Zn^+2 + Cu Ecell = 0.515

Therefore, since Ecell is 0.515 v. then the Zn==> Zn^+2 + 2e must be 0.515 v; that is, the ?? marks above will be 0.515 v. The others are done in a similar manner and I think your listings are correct.
Then the instructions are to list them as reduction reactions in order of decreasing reduction potential. So the most negative goes at the bottom and the most positive goes at the top. Written as reductions, they are
Zn^+2 + 2e ==> Zn -0.515
Al^+3 + 3e ==> Al -0.729
Mg^+2 + 2e ==> Mg -1.728

I think this takes care of all of your follow up questions.

1. 👍 0
2. 👎 0
5. vjdc mfvcro fbdvtzqnm cnpxkvhgo tdbmc bgjmx cskfb

1. 👍 0
2. 👎 0

1. 👍 0
2. 👎 0
7. aa aer you all bad at chem mannn

1. 👍 0
2. 👎 0
posted by Jimmy
8. That's rude, Dr.Bob.

1. 👍 0
2. 👎 0
9. ^ Sorry Dr.Bob222 on behalf of these people from UCI

1. 👍 0
2. 👎 0
posted by anon
10. but how do you titrate an analyte without a buffer solution reacting with thermonuclear hygiene?

1. 👍 0
2. 👎 0
11. Hi, I have this similar lab just with different values and I Completely understand it thanks to your steps, however at the end when the voltages are suddenly turned to negative i am confused. Why are they negative all of a sudden?

1. 👍 0
2. 👎 0
posted by Cindy

## Similar Questions

1. ### chemistry- DrBob

Last week i did a lab and i just wanted to show you my results because my computer was acting weird when i used the voltage probe.(I just needed you to check that the cathode and anodes and voltages make sense) CuSO4(cathode)and

asked by Amy on May 16, 2008
2. ### chemistry- DrBob

summarize part a ( the data i showed you) in a data table containing heading: anode, cathode, overall cell reaction, and cell potential . for each cell combination, write the anode half reaction, the cathode half reaction, the

asked by amy on May 16, 2008
3. ### AP Chemistry

For the reaction ? Al + ? CuSO4 ↽⇀? Al2(SO4)3+? Cu a maximum of how many moles of Al2(SO4)3 could be formed from 5.95 mol of Al and 1.31 mol of CuSO4? Answer in units of mol.

asked by Matt on September 19, 2012
4. ### AP Chemistry

For the reaction ? Al + ? CuSO4 ↽⇀? Al2(SO4)3+? Cu a maximum of how many moles of Al2(SO4)3 could be formed from 5.49 mol of Al and 4.09 mol of CuSO4? Answer in units of mol

asked by Tina on September 13, 2011
5. ### chemistry

Suggest chemical reactions that could be used to identify the contents of four unlabeled bottles containing solutions of the following substances: CuSO4, Al2(SO4)3, Pb(NO3)2, and FeSO4.

asked by ABCD on July 3, 2011
6. ### Chemistry

so if the equation I'm using is 3CuSO4+2AL(NO3)3 arrow 3CU(NO3)2+Al2(SO4)3 how many moles of CuSO4 would be needed to produce 8 moles of Al2(SO4)3? Would the answer be 8?

asked by Marc on April 21, 2009
7. ### Chemistry

Hi!! pls i need help! 1) Mg + Al2(So4)3 2)Fe + CuSO4 3)Mg + CuSO4 4)Zn + CuSO4 5)Mg + FeSO4 FOR ALL 5: a)balanced eq b)ionic equation c) net ionic PLEASE HELP ME..THERE ARE SOO MANY MORE QUESTIONS I NEED TO DO!!!:(

asked by Deepa on February 18, 2014
8. ### chemistry

Hi, the question is: in each reaction, identify what have been oxidised and reduced. The equation is: 2Al + 3H2SO4 ---> Al2(SO4)3 + 3H2 I am confused about the Al2(SO4)3 I know what oxides and reductions us I just don't understant

asked by Franky on October 6, 2011
9. ### CHEMISTRY

Hi!! pls i need help! 1) Mg + Al2(So4)3 2)Fe + CuSO4 3)Mg + CuSO4 4)Zn + CuSO4 5)Mg + FeSO4 FOR ALL 5: a)balanced eq b)ionic equation c) net ionic

asked by Deepa on February 18, 2014
10. ### chemistry 110

what combinations of Zn, Al, ZnSO4 and Al2(SO4)3 would you make to illustrate that Zn is less active than Al?

asked by shannon on October 28, 2010

More Similar Questions