# Any chemistry lovers?

A voltaic cell consists of Mn/Mn2+ and Cd/Cd2+ half-cells with the following initial concentrations: [Mn2+] = 0.090 M; [Cd2+] = 0.060 M.
A)What is the intital Ecell?
B)What is Ecell when [Cd2+} reaches .050M.
C) What is [Mn2+] when Ecell reaches 0.055V?
D) What are the equilibirum concentrations of the ions?

I only need to know part C and D. How is it done, the answers are:
C) 0.150M and D)[Mn2+]= 0.150M and [Cd2+]=6.4X10^(-28)M

Relevant data: Mn(S) to its ion is 1.18V and for Cd ion to solid, it is -.40V, the total is .78V.

Thanks in advance.

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asked by stumped
1. fiyu

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posted by vjk
2. C)
Set up the equation with known information:

E cell = E naught -(0.0592/n)*log Q

0.055=0.78-(0.0592/2)*log Q

SOLVE for Q so that you will be able to equate Q to the concentration of Mn2+ over the concentration of Cd2+ (according to chemical equilibria rules):

-0.725 = -(0.0592/2)*log Q

log Q = 24.493

Q = 3.11*10^(24)

NOW set equal to [Mn2+]/[Cd2+] with concentrations from question:

3.11*10^(24) = (0.090+x)/(0.060-x)

SOLVE

[1.868*10^(23)]- (3.11*10^(24))x = 0.09+x

the 0.09 has barely any affect (same with the +1x), so:

1.868*10^(23) = [3.11*10^(24)]x

x= 0.06

ADD to original concentration of Mn2+ (=0.090):

0.06 + 0.090 = 0.150M = [Mn2+]

PART D)
At equilibrium, Ecell= 0. Why? Because the term (0.0592V/n)*log Q becomes so big that it equals E naught which means that E cell = 0.

So, we set E cell equal to zero.

0 = E naught - (0.0592/n)*log Q

Now, fill in what we know:

0 = 0.78-(0.0592/2)*log((0.090+x)/(0.060-x))

Then we find through the same methods in part C that since [Mn2+]= 0.150M, that [Cd2+]= 6.68*10^(-28)M

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posted by Megan

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