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What is the cell potential when the concentration of Ni^2+ has fallen to 0.500 M and other questions

Submitted by RSS on Sun, 04/17/2011 - 17:34

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?


Zn  +  Ni+2  -->  Zn+2  +  Ni

So the 1/2 reactions and standard 1/2 cell potentials are
          oxid.    Zn  -->  Zn+2  +  2e-      Eo oxid =+ .7628
          red/      Ni+2  +  2e-  -->  Ni      Eo red  = -.23
Standard Cell potential is +.5368 volts

Since the initial concentration of the Ni+2 was 1.30 M and we are asked to find the cell potential when it has been reduced to .500M, a change of .800M.
That means that the concentration of the Zn+2 must have been increased by the same amount from .100M to .900M.

Use the Nernst Equation to find Ecell where Q = [Zn+2]/[Ni+2] =  .900M/.500M
Ecell = E0cell - (RT/nF)lnQ

For the next set I assume that you have a typo and the cell potential is .45 Volts (not M).  In this case, since you know the standard and non-standard cell potentials, you can solve the Nernst equation and determine the value for Q.

but you if you let x be the change in concentration then [Zn+2] = .100 + x
and [Ni+2] = 1.30 - x.

Substitute in the equation where Q = [Zn+2]/[Ni+2] and determine the value of x and then find the value of the concentrations of zinc and nickel ions.