The voltage generated by the zinc concentration cell described by,
Zn(s)|Zn2 (aq, 0.100 M)||Zn2 (aq, ? M)|Zn(s)
is 24.0 mV at 25 °C. Calculate the concentration of the Zn2 (aq) ion at the cathode.
How would I go about this problem? I am extremely confused :/
Thank you so much in advance!!!
Ecell = (-0.0592/2)log Q
You know Ecell is 0.024 v
Q from the equation you wrote is
(?Zn^2+)/(Zn^2+ 0.1) or it may be better to write Q as
Q = (x/0.1) and solve for x = ?M Zn^2+ on the right hand side of the equation. Post your work if you get stuck.
To solve this problem, you need to use the Nernst equation, which relates the voltage (Ecell) of an electrochemical cell to the concentrations (or activities) of the reactants and products involved. The Nernst equation is given by:
Ecell = E°cell - (RT / nF) * ln(Q)
where:
- Ecell is the cell potential
- E°cell is the standard cell potential
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin (25 °C = 298 K)
- n is the number of electrons transferred in the cell reaction
- F is Faraday's constant (96,485 C/mol)
- ln is the natural logarithm
- Q is the reaction quotient
In this case, you have a concentration cell with zinc electrodes and zinc ions (Zn2+) in both compartments. The reaction involved is the reduction of Zn2+ to Zn:
Zn2+ (aq) + 2e- → Zn (s)
Given the information in the problem, we can assume that the standard cell potential (E°cell) is 0 V because both half-cells have the same species. Therefore, the Nernst equation simplifies to:
Ecell = - (RT / nF) * ln(Q)
We can rearrange the equation to solve for Q:
Q = e^(-(Ecell * n * F) / (RT))
Substituting the known values:
- Ecell = -0.024 V (convert mV to V)
- n = 2 (according to the balanced equation)
- R = 8.314 J/(mol·K)
- T = 298 K
- F = 96,485 C/mol
Q = e^(-((-0.024 V) * 2 * (96,485 C/mol))/(8.314 J/(mol·K) * 298 K))
Simplifying the equation gives:
Q = e^(0.128)
Now, Q represents the concentration ratio of the zinc ions at the cathode (Zn2+(aq)) to those at the anode (Zn2+(aq)).
Since the cathode concentration is the unknown, let's represent it as "x".
Therefore, Q = (x M) / (0.100 M), where 0.100 M is the concentration of Zn2+(aq) at the anode.
We can now solve for x by equating Q to the calculated value:
(e^(0.128)) = x / (0.100)
x = (e^(0.128)) * (0.100)
Using a calculator, calculate the value of (e^(0.128)) * (0.100) to find the concentration of Zn2+(aq) at the cathode.
Remember to convert the final answer to the appropriate units (Molarity in this case).