U9.9 · Cell Potential Under Nonstandard ConditionsPractice · 非标准条件下的电池电势
Lessons covered:U9.9 Cell Potential Under Nonstandard Conditions
- 1Multiple choice
A galvanic cell running Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) starts at standard conditions (E° = 0.79 V). More solid CuSO₄ is dissolved on the copper side, raising [Cu²⁺]. How does the cell potential E change?
ExplanationCu²⁺ is a reactant, so raising its concentration lowers Q and increases the drive toward products: E increases. By the Nernst equation E = E° − (0.0592/n) log Q, Q < 1 gives log Q < 0 and E > E°. E° is the standard-state value and doesn’t change with concentration; E does.
- 2Numeric
At 25 °C, Fe(s) + Cu²⁺(aq) → Fe²⁺(aq) + Cu(s) has E° = 0.79 V. If [Fe²⁺] = 1.0 M and [Cu²⁺] = 0.010 M, what is the cell potential E? (E = E° − (0.0592/n) log Q)
VExplanationQ = [Fe²⁺] ÷ [Cu²⁺] = 1.0 ÷ 0.010 = 100 and n = 2. E = 0.79 − (0.0592/2) × log 100 = 0.79 − 0.0296 × 2 = 0.73 V. With less reactant Cu²⁺, E is a little below E°.
- 3Multiple choice
A galvanic cell starts at standard conditions and discharges until it can no longer supply current. How do Q and E change along the way?
ExplanationAs the cell discharges, reactants are used up and products build up, so Q rises and E = E° − (RT/nF) ln Q gradually falls. When Q = K the system is at equilibrium, E = 0, and the battery is dead.
- 4Numeric
Two half-cells each have a nickel strip in NiSO₄ solution, one at [Ni²⁺] = 1.0 M and the other at 0.0010 M, joined by a salt bridge. What is the potential of this concentration cell at 25 °C? (E = E° − (0.0592/n) log Q)
VExplanationThe electrodes are identical, so E° = 0. Ni is oxidized on the dilute side (anode) and Ni²⁺ is reduced on the concentrated side (cathode); overall Ni²⁺(1.0 M) → Ni²⁺(0.0010 M), so Q = 0.0010 ÷ 1.0 = 1.0 × 10⁻³ and n = 2. E = 0 − (0.0592/2) × log(1.0 × 10⁻³) = 0.0296 × 3 = 0.089 V. When the two concentrations become equal, E = 0.
- 5Multiple choice
In a galvanic cell running Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), [Zn²⁺] = 0.010 M and [Ag⁺] = 1.0 M. How does E compare with E°?
ExplanationQ = [Zn²⁺] ÷ [Ag⁺]² = 0.010 ÷ 1.0² = 0.010 < 1. With little product Zn²⁺, the drive toward products is stronger, so E > E° (by about 0.06 V). If any species isn’t at 1 M, the cell isn’t at standard conditions.
- 6Multiple choice
The graph shows the cell potential of at 25 °C against log Q, where Q = [Zn²⁺]/[Cu²⁺]. When [Zn²⁺] is 100 times [Cu²⁺], what is Ecell?
ExplanationQ = 100, so log Q = 2. Read the line at x = 2: about 1.04 V. The line is E = E° − (0.0592/n) log Q: at log Q = 0 (equal ion concentrations) E = E° = 1.10 V, and the slope is −0.0592/2 ≈ −0.030 V. More product (Zn²⁺) and less reactant (Cu²⁺) lowers the potential; choosing 1.16 V uses the reciprocal of Q (log Q = −2).
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