I need help with this question. What should be the cell potential if the [Cu2+] is 1.0 x 10-5 mol/L?
please help
I suppose you mean the half cell potential.
E = Eo - (0.0592/n)Log(red/ox)
Cu^+2 + 2e ==> Cu
Cu is reduced form.
Cu^+2 is oxidized form.
Plug into the above equation and solve for E. This is E for the concn given as well as for T = 25 C.
42...
It's always 42
To determine the cell potential, you need to use the Nernst equation, which relates the cell potential to the concentration of the species involved in the electrochemical reaction.
The Nernst equation is given as follows:
Ecell = E°cell - (0.0592/n) * log(Q)
Where:
- Ecell represents the cell potential
- E°cell represents the standard cell potential
- n represents the number of electrons transferred in the Electrochemical reaction
- Q represents the reaction quotient, which is calculated using the concentrations of the species involved in the electrochemical reaction
In this specific case, you have not provided the entire reaction, so I will assume it involves copper ions (Cu2+). Therefore, let's assume the reaction to be:
Cu(s) | Cu2+(aq) || Cu2+(aq) | Cu(s)
Since the reaction involves the same species on both sides of the cell, the cell potential is zero, and the concentration of [Cu2+] does not affect the cell potential. Therefore, the cell potential would be zero, regardless of the concentration of [Cu2+].