I'm having trouble with the following question, any help would be greatly appreciated.
A voltaic cell consists of a Zn/Zn^2+ half-cell and a Ni/Ni^2+ half-cell at 25 C . The initial concentrations of Ni^2+ and Zn^2+ are 1.30 M and 0.100 M , respectively.
What is the cell potential when the concentration of Ni^2+ has fallen to 0.500 M?
What is the concentrations of Ni^2+ when the cell potential falls to 0.45 M?
What is the concentration of Zn^2+ when the cell potential falls to 0.45 M?
3 answers
I haven't tried to work on this problem at all because I don't understand how Ni2+ ion can start at 0.100M, use some of it and END UP FALLING to 0.500M. It appears to me that Ni2+ can become smaller but not larger.
The second part of this question does not make sense cell potential has to be in V not M...M is for concentration
I got the correct answer by making a simple Initial/Final chart. the initial [Ni^2+] = 1.30, the final [Ni^2+] = 0.500. The change is 0.80 M. This means you would add 0.80 to the initial [Zn^2+] (which is .100 M) and you would get a final [Zn^2+] of 0.90 M.
The next step is to find the new Q which is [Zn^2+]/[Ni^2+] = (0.90)/(0.500). Plug this in the Nernst equation and you will find the cell potential.
The next step is to find the new Q which is [Zn^2+]/[Ni^2+] = (0.90)/(0.500). Plug this in the Nernst equation and you will find the cell potential.
0 answers