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Cell potential and E cell calculations using standard reduction potential series

Submitted by Ohso on Sat, 04/11/2009 - 23:56

Hi, I am having some issues solving for these problems. Any help would be good, thanks.

The first question is...

Is H+(aq) capable of oxidizing Cu(s) to Cu2+(aq)?
What is E°cell in this case?

I know the answer is No however I do not know how to calculate the Eo cell.

The second question...

An electrochemical (galvanic) cell consists of a nickel metal electrode immersed in a solution with [Ni2+] = 1.0 M separated by a porous disk from an aluminum metal electrode.

(a) What is the potential of this cell at 25°C if the aluminum electrode is placed in a solution in which [Al3+] = 5.4  10-3 M?

(b) When the aluminum electrode is placed in a certain solution in which [Al3+] is unknown, the measured cell potential at 25°C is 1.55 V. Calculate [Al3+] in the unknown solution. (Assume Al is oxidized.)

Write the two reduction half equations using your data - something like:
2H+(aq) + 2e- -----> H2(g)    Eo = 0.00V
Cu2+(aq) + 2e- -----> Cu(s)  Eo =  0.34V
The values show that the lower reaction is more likely as a reduction and will probably cause the reaction above it to go in reverse - as an oxidation.
In your example, you wish to find if Cu can be oxidised to Cu2+

So reverse this equation:
2H+(aq) + 2e- -----> H2(g)    Eo = 0.00V
Cu(s) -----> Cu2+(aq) + 2e-  -Eo = -0.34V
As there are the same number of electrons in both reactions, the equations can be added. The Eo values are added to give the emf of the cell (under standard conditions)
2H+(aq) + Cu(s) -----> H2(g) + Cu2+(aq) emf = -0.34V
As this is a negative value, it will not work.

The reactions which work are:
H2(g) + Cu2+(aq) -----> 2H+(aq) + Cu(s) emf = 0.34V

As a general rule, reactions lower in the Eo table will go as written (reduction) and cause reactions above to go in reverse (as oxidation).

thanks but that was my original answer and I inputed it online and got it wrong.

There is nothing wrong with the method. Are you sure you are using the correct values for Eo as some data tables give slightly different/more accurate values(?).

Ok I was trying 0.34 V t is actually -0.34 V