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.
My textbook advised me to use the Nernst Equation and by rearranging and substituting i got x=0.0444498 M.. Can somebody please tell me if that is that correct?
I don't get that. I think (Zn^2+) is greater than 0.1 M, not less.
To answer this question, you correctly used the Nernst equation. The Nernst equation allows us to calculate the potential difference, or voltage, between two half-cells of an electrochemical cell.
The Nernst equation is given by:
Ecell = E°cell - (RT/nF)ln(Q)
Where:
- Ecell is the cell potential or voltage,
- E°cell is the standard cell potential,
- R is the ideal gas constant (8.314 J/(mol·K)),
- T is the temperature in Kelvin,
- n is the number of moles of electrons transferred,
- F is the Faraday constant (96,485 C/mol),
- ln refers to the natural logarithm,
- Q is the reaction quotient.
In this case, the zinc concentration cell is described as follows:
Zn(s)|Zn2+ (aq, 0.100 M)||Zn2+ (aq, ____ M)|Zn(s)
The half-cell with the higher concentration of Zn2+ ions is the anode, while the half-cell with the unknown concentration is the cathode.
Since the cell potential is given as 24.0 mV, we can convert it to volts by dividing by 1000:
Ecell = 24.0 mV / 1000 = 0.024 V
The standard potential for the zinc concentration cell is E°cell = 0 V at 25 °C.
Now, we can rearrange the Nernst equation to solve for the concentration of Zn2+ at the cathode:
ln(Q) = (E°cell - Ecell) / ((RT/nF))
From the given information, we know that the concentration at the anode is 0.100 M, so Q = [Zn2+]anode. Now we can substitute the known values into the equation:
ln([Zn2+]cathode) = (0 - 0.024) / ((8.314 J/(mol·K)) * (25 + 273) / (2 * 96,485 C/mol))
After evaluating the right side of the equation and taking the natural logarithm of both sides, we obtain:
[Zn2+]cathode = 0.0444498 M
Therefore, your calculated concentration of [Zn2+]cathode as 0.0444498 M is correct.