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Electrochemical Cell questions involving Mg and Cu electrodes in orange drink

Submitted by Anonymous (not verified) on Sun, 04/06/2008 - 18:03

Ok, I'm trying to figure out some questions regarding an electrochemical cell for and orange juice clock involving a magnesium strip and a copper strip in 400ml of orange juice with wireing connecting them to a clock.

So far I have determined that the Mg is what is being oxidized (and is hence the anode) 1)but I'm a little stuck on what is being reduced - I know it can't be the copper because there are no copper ions in the OJ so what else? Water maybe?

1/2 reactions:
oxidized:  Mg(s) -> Mg2+(aq) + 2e-  whereE0=2.37V (vs. SHE)
reduced:  2H+(aq) + 2e- -> H2(g) ?where E0=0V (vs. SHE)
overall:  Mg(s) + 2H+(aq) -> Mg2+(aq) +H2(g)  ?

If so, the E0cell would be
E0cell=E0(reduced 1/2 cell) - E0(oxidized 1/2 cell)= (0V)-(2.37)V
E0cell=2.37V

2. The pH of the orange juice was measured to be 1.5. What is the measurement electrochemical potential of the organic juice clock electrochemical cell?

3.The clock requires a minimum voltage of 1.5 volts for it to function. Now, if we replace the magnesium electrode with a lead electrode, will the clock still run? How about replacing the electrode with a stainless steel plate?

4.Would the system work if we put Cu2+ ions into the orange juice solution?

5. What would happen if we titrate the acid solution with a strong base while the clock is running? Explain your observation.

Thanks in advance!!!!


Agree that the Eo for the reaction is 2.37,

You would need to use the Nernst equation to come up with a non-standard solution:
            E = Eo -.059/2  log Q where Q = [Mg+2] PH2 / [H+]

Although we can calculate the value of [H+] from the pH, without values for pH2 and the [Mg+2], I don't see how you can calculate an answer.

Using a Pb strip would only generate a overall cell voltage of .13, so it wouldn't work.  Don't know what the overall cell voltage would be for steel (Fe?), but don't think it would work since pure Fe wouldn't work.

Titrating the acid would reduce the H+ ions, this should reduce the overall cell voltage, the clock would stop running at some point.

Adding Cu+2 to the orange juice should increase the overall cell voltage since Cu+2 is a better oxidizing agent than H+.


spock wrote:

Agree that the Eo for the reaction is 2.37,

You would need to use the Nernst equation to come up with a non-standard solution:
            E = Eo -.059/2  log Q where Q = [Mg+2] PH2 / [H+]

Although we can calculate the value of [H+] from the pH, without values for pH2 and the [Mg+2], I don't see how you can calculate an answer.

Using the Nernst equation, can't you derive E=E0cell + .0591(pH outside-pHinside)
where you can disregard the pH inside becasuse the pH of the inside of electrode is constant so if you knew the pH you could still solve for Ecell?

Also, what happens to the voltage if we put Mg2+ into the system?
-Are the complex ions of Mg2+ with citric acid important to the potential value?


Perhaps you can derive E=E0cell + .0591(pH outside-pHinside) but I've never seen that equation before and I'm not sure what you would mean by pH inside and pH outside of the cell.

I'm not aware of complex ions forming between citrate ions and Mg+2 ions.  Normally complex ions involve transition metal ions.

Are these questions for a general chemistry course or physical chem course? 


General Chem
I thinks it's the pH of the inside and ouside of the electrode...