Problem
Consider the galvanic cell \(\mathrm{Zn(s)\,|\,Zn^{2+}(0.010\,M)\,||\,Cu^{2+}(1.00\,M)\,|\,Cu(s)}\) at \(25^\circ\mathrm{C}\). Using the nernst equation, determine the cell potential \(E\) given \(E^\circ(\mathrm{Cu^{2+}/Cu})=+0.34\,\mathrm{V}\) and \(E^\circ(\mathrm{Zn^{2+}/Zn})=-0.76\,\mathrm{V}\).
What “nonstandard” means: concentrations are not always \(1.00\,\mathrm{M}\), thus \(E \neq E^\circ\).
Concentration Effect: Since \([\mathrm{Zn^{2+}}]\) is low, Le Chatelier’s principle suggests a higher driving force for oxidation, increasing the cell potential.
Solution
1) Write the overall redox reaction and identify \(n\)
The half-reactions for this cell are:
- Anode (Oxidation): \(\mathrm{Zn(s) \rightarrow Zn^{2+}(aq) + 2e^-}\)
- Cathode (Reduction): \(\mathrm{Cu^{2+}(aq) + 2e^- \rightarrow Cu(s)}\)
Adding these gives the net reaction:
The number of electrons transferred is \(n=2\).
2) Compute the standard cell potential \(E^\circ_{\text{cell}}\)
Use \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\):
3) Determine the reaction quotient \(Q\)
Solids (Zn and Cu) are excluded from the mass-action expression:
Calculating the base-10 logarithm:
4) Apply the Nernst Equation
At \(25^\circ\mathrm{C}\) (standard lab temperature), we use the simplified form:
Inserting the values:
Result: \(E \approx 1.16\,\mathrm{V}\). The potential is increased by roughly \(60\,\mathrm{mV}\) due to the low zinc concentration.
Visualization: Nernst Equation Dashboard
Summary Table
| Quantity | Meaning | Value |
|---|---|---|
| \(E^\circ_{\text{cell}}\) | Standard cell potential | \(1.10\,\mathrm{V}\) |
| \(n\) | Electrons transferred | \(2\) |
| \(Q\) | Reaction quotient | \(0.010\) |
| \(\log_{10}(Q)\) | Base-10 logarithm of \(Q\) | \(-2.00\) |
| \(E\) | Nonstandard potential | \(\approx 1.16\,\mathrm{V}\) |
Final Answer
Using the Nernst equation at \(25^\circ\mathrm{C}\) with \(E^\circ_{\text{cell}}=1.10\,\mathrm{V}\), \(n=2\), and \(Q=\frac{0.010}{1.00}=0.010\), the resulting cell potential is \(E \approx 1.16\,\mathrm{V}\).