Calculate concentration of halfcell

In summary, the conversation discusses the use of the Nernst equation to calculate the concentration of Cd2+ in a voltaic cell at 25oC with an initial cell potential of +0.768 V. The [Mn2+] concentration is known to be 0.500 M and the given half-cell reactions are Cd+2(aq) + 2e- = Cd(s) and Mn+2(aq) + 2e- = Mn(s). The possible options for the [Cd2+] concentration are listed and the conversation concludes with the problem being solved.
  • #1
m0286
63
0
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
 
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  • #2
got it!

thnx for lookin but we figured it out :)
 
  • #3


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.
 

1. What is meant by the concentration of a halfcell?

The concentration of a halfcell refers to the amount of a particular substance, typically ions, present in one half of an electrochemical cell. It is an important factor in determining the potential difference and overall functioning of the cell.

2. How do you calculate the concentration of a halfcell?

The concentration of a halfcell can be calculated by using the Nernst equation, which takes into account the standard reduction potential, temperature, and concentration of the ions in the halfcell. The equation is E = E° - (0.0592/n) x log([ion]n+ / [ion]n-), where E is the cell potential, E° is the standard reduction potential, n is the number of electrons involved in the reaction, and [ion]n+ and [ion]n- are the concentrations of the oxidized and reduced forms of the ion, respectively.

3. Why is it important to know the concentration of a halfcell?

The concentration of a halfcell is important because it affects the cell potential and overall performance of the electrochemical cell. A higher concentration of ions can lead to a higher cell potential and greater efficiency, while a lower concentration can result in a lower potential and reduced functionality.

4. What factors can affect the concentration of a halfcell?

The concentration of a halfcell can be affected by various factors such as temperature, pressure, and the presence of other substances. Changes in these factors can result in a shift in the equilibrium of the cell reaction, leading to a change in the concentration of ions and ultimately altering the cell potential.

5. How can the concentration of a halfcell be controlled?

The concentration of a halfcell can be controlled by adjusting the amount of the substance present in the halfcell or by changing the temperature and pressure conditions. In some cases, a concentration gradient can be created by using a semipermeable membrane or by actively pumping ions into or out of the halfcell to maintain a desired concentration level.

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