A Cu/Cu2+ concentration cell has a voltage of 0.35 V at 25°C. The concentration of Cu2+ in one of the half-cells is 1.76 × 10−3 What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell is the lower of the two concentrations.)

Question

BlackTom AI · Accepted Answer

For a concentration cell the Nernst equation gives E = (RT/nF) ln([Cu2+]high/[Cu2+]low).

1) Relevant concepts
- A concentration cell uses the same redox couple on both electrodes; the cell potential arises from the concentration ratio of the ions.
- Nernst equation (at T = 25.0 °C = 298 K): E = (RT/nF) ln(Q). For electron transfer n = 2 for Cu2+ + 2 e− ⇌ Cu(s).
- Commonly used forms: E = (0.05916 V/n) log10([oxidant]high/[oxidant]low) at 25 °C.

2) Step-by-step reasoning
- Given: E = 0.35 V, T = 298 K, known [Cu2+] = 1.76 × 10−3 M, unknown is the lower concentration, and n = 2.
- Use the base-10 form: E = (0.05916/2) log10([high]/[low]).

3) Calculations (all intermediate steps)
- (0.05916/2) = 0.02958 V
- Rearranged: log10([high]/[low]) = E / 0.02958 = 0.35 / 0.02958 ≈ 11.835
- Ratio [high]/[low] = 10^11.835 ≈ 6.83 × 10^11
- If the known concentration 1.76 × 10−3 M is the higher concentration, then
  [low] = (1.76 × 10−3) / (6.83 × 10^11) ≈ 2.62 × 10−15 M (after appropriate rounding).

4) Why the correct answer is correct
- The Nernst equation directly links the observed cell voltage to the logarithm of the concentration ratio. Solving for the lower concentration by dividing the known higher concentration by the large ratio produced by 0.35 V yields the extremely small value ≈ 2.62 × 10−15 M. This is consistent with the assumption that the unknown is the lower concentration.

5) Why other options are incorrect
- 1.49 × 10−12 M is too large by about three orders of magnitude compared with the ratio required by 0.35 V.
- 0.44 M and 0.50 M are orders of magnitude larger than the given 1.76 × 10−3 M and contradict the statement that the unknown is the lower concentration.
- −11.8 M is physically impossible (negative concentration).

6) Conclusion
- The calculated lower concentration of Cu2+ is 2.62 × 10−15 M, which matches the provided correct answer.

A Cu/Cu2+ concentration cell has a voltage of 0.35 V at 25°C. The concentration of Cu2+ in one of the half-cells is 1.76 × 10−3 What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell is the lower of the two concentrations.)单项选择题

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