Calculate concentration of halfcell

[m0286]

Hello
Im completely stuck on this question and don't know where to begin or what to do.. and help or guidance is appreciated:

A voltaic cell at 25oC consists of Mn/Mn2+ and Cd/Cd2+ half-cells with an initial cell potential of +0.768 V. The [Mn2+] concentration is 0.500 M. Use the Nernst equation to calculate the [Cd2+] concentration in the Cd/Cd2+ half-cell.
Cd+2(aq) + 2e- = Cd(s) . . . . . Eo = -0.40 V

Mn+2(aq) + 2e- = Mn(s) . . . . . Eo = -1.18 V

a. 0.010 M
b. 0.050 M
c. 0.20 M
d. 0.5
e. 0.25
its an online assignment and its due at 6am so any help soon is appreciated THANKS

 

[m0286]

got it!

thnx for lookin but we figured it out :)

 

[Blueshift5]

I would first start by writing out the balanced half-cell reactions for both the Mn/Mn2+ and Cd/Cd2+ half-cells:

Mn+2(aq) + 2e- = Mn(s) (Eo = -1.18 V)
Cd+2(aq) + 2e- = Cd(s) (Eo = -0.40 V)

Next, I would use the Nernst equation to calculate the cell potential at 25oC:

Ecell = Eo - (0.0592/n)log(Q)
Where:
Ecell = cell potential (in volts)
Eo = standard cell potential (in volts)
n = number of electrons transferred in the half-cell reactions (in this case, n = 2)
Q = reaction quotient (in this case, [Mn2+]/[Cd2+])

We know that the initial cell potential is +0.768 V, so we can plug that in for Ecell. We also know the standard cell potentials for both half-cells, so we can plug those in as well. The only unknown in the equation is Q, which we can solve for using the given [Mn2+] concentration (0.500 M) and the unknown [Cd2+] concentration:

Ecell = 0.768 V = (-1.18 V) - (0.0592/2)log([Mn2+]/[Cd2+])
0.768 V = -1.18 V - (0.0296)log([0.500 M]/[Cd2+])
1.948 V = log([0.500 M]/[Cd2+])
10^1.948 = [0.500 M]/[Cd2+]
88.678 = [0.500 M]/[Cd2+]
88.678 x [Cd2+] = 0.500 M
[Cd2+] = 0.500 M / 88.678
[Cd2+] = 0.00564 M

Therefore, the [Cd2+] concentration in the Cd/Cd2+ half-cell is 0.00564 M or 5.64 x 10^-3 M.

In response to the given answer choices, the correct answer would be option a. 0.010 M.