Chapter 17 Source Answers and Examples

Example Problem 9.1.1 — AP edition reviewed replacement.

Reviewed circuit enrichment: four ideal 1.50 V cells are connected in series in the same polarity across a 24.0 Ω load. Each cell has zero internal resistance. Find the load voltage and current. This complete task replaces the underspecified source drawing and fixed-current rule.

Series voltages add in the stipulated polarity: V=4×1.50=6.00 V.

The load determines current: I=V/R=6.00/24.0=0.250 A. The same current passes through all four series cells.

This ideal enrichment model explicitly specifies the load and internal resistance. A battery’s marked current alone cannot determine circuit current.

Example Problem 9.1.2 — AP edition reviewed replacement.

Reviewed circuit enrichment: two identical ideal 3.00 V cells are connected in parallel with like-polarity terminals together across a 12.0 Ω load. Specify equal current sharing, zero internal resistance and no other loads. Find total current and each cell’s current.

Parallel identical ideal sources keep the terminal voltage at 3.00 V. The two voltages do not add.

I(total)=V/R=3.00/12.0=0.250 A. Under the stipulated equal-sharing model, each cell supplies 0.125 A.

Real sharing depends on source voltages and internal resistances. The source’s 4.5 V at 2 A result does not solve this newly authored task.

Example Problem 9.2.1 — AP edition reviewed replacement.

Reviewed capacity task: one ideal cell has 600 mA h of usable charge capacity and supplies constant current 200 mA. Determine its lifetime. The original multi-battery circuit is not part of this replacement.

Keep charge capacity and current distinct: 600 mA h is a charge, not a current.

t=Q/I=(600 mA h)/(200 mA)=3.00 h. Alternatively Q=0.600 A h×3600 s/h=2160 C and t=2160/0.200=10800 s.

This answer assumes the stated usable capacity and constant current. It predicts no current in an unspecified multi-cell circuit.

Example Problem 9.2.2 — AP edition reviewed replacement.

Reviewed Faraday task: one ideal Zn/Cu cell has 5.00 g of usable Zn, excess Cu²⁺ and constant current 0.500 A. Assume Zn→Zn²⁺+2 e⁻, 100% utilization/current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Find lifetime.

n(Zn)=5.00/65.38=0.0764760 mol. Each zinc atom supplies two electrons, so n(e⁻)=0.152952 mol.

Q=n(e⁻)F=(5.00/65.38)×2×96485.3=14757.6 C.

t=Q/I=14757.6/0.500=29515.2 s=8.19868 h≈8.20 h. This solves a single-cell capacity model; it makes no prediction about the source’s unspecified circuit.

Example Problem 9.2.3 — AP edition reviewed replacement.

Reviewed ideal sacrificial-anode capacity task: a 5.00 kg Zn anode has 80.0% usable mass and supplies constant current 0.0500 A. Assume Zn→Zn²⁺+2 e⁻, 100% useful-current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Estimate the capacity-limited lifetime.

Usable Zn mass=5000×0.800=4000 g. n(e⁻)=2×4000/65.38.

Q=(4000/65.38)×2F=1.18061×10⁷ C. t=Q/0.0500=2.36122×10⁸ s.

t≈2732.89 days≈7.48 years (365.25 days per year). This is an ideal charge-capacity estimate, not an engineering protection-life guarantee. The original unrelated circuit solution is superseded.

Example Problem 9.2.4 — AP edition reviewed replacement.

Reviewed ideal capacity task: a 10.0 m long cylindrical pipe segment has diameter 5.00 m and stipulated uniform current density 0.300 A m⁻² over its lateral area. Its Zn anode has 10.0 kg usable mass. Assume Zn→Zn²⁺+2 e⁻, 100% utilization/current efficiency, M(Zn)=65.38 g mol⁻¹ and F=96485.3 C mol⁻¹. Estimate lifetime.

Use the stipulated segment’s lateral area: A=πDL=π×5.00×10.0=157.080 m². I=jA=0.300×157.080=47.1239 A.

Q=(10000/65.38)×2F=2.95152×10⁷ C.

t=Q/I≈626333 s≈7.24922 days≈7.25 days (three significant figures). This is a stated ideal segment model, not a design assessment for the original 50 km pipeline. Its solution must not be replaced by a boat-anode or single-cell example.

Example Problem 9.3.1 — AP edition reviewed replacement.

Reviewed explicit gas convention: ideal water electrolysis passes 4.00 A for 6.00 s with 100% Faradaic efficiency for H₂ formation. Find the ideal H₂ volume at 273.15 K and 100.0 kPa. Use F=96485.3 C mol⁻¹ and R=8.314462618 L kPa mol⁻¹ K⁻¹.

Q=It=4.00×6.00=24.0 C. Two moles of electrons form one mole of H₂, so n(H₂)=24.0/(2F)=1.24371×10⁻⁴ mol.

V=nRT/P=[24.0/(2×96485.3)]×8.314462618×273.15/100.0=0.00282459 L≈2.82 mL.

The temperature and pressure are stated explicitly; a different STP pressure convention gives a different molar volume. The source gas calculation is preserved in provenance.

Example Problem 9.3.2 — AP edition reviewed replacement.

Reviewed chloride-electrolysis model: stipulate that an aqueous chloride cell produces H₂ at the cathode and Cl₂ at the anode, each with 100% Faradaic efficiency. Current is constant at 5.00 A for 1.00 h. Find the ideal volume of each separately collected gas at 273.15 K and 100.0 kPa. Use F=96485.3 C mol⁻¹ and R=8.314462618 L kPa mol⁻¹ K⁻¹. This task does not infer an operating-voltage window from saturated NaCl.

The stipulated half-reactions are 2 H₂O+2 e⁻→H₂+2 OH⁻ and 2 Cl⁻→Cl₂+2 e⁻. They give 2 H₂O+2 Cl⁻→H₂+Cl₂+2 OH⁻, conserving atoms and charge.

Q=It=5.00×3600=18000 C. Each gas requires two moles of electrons per mole, so n(H₂)=n(Cl₂)=18000/(2F)=0.0932785 mol.

For either gas, V=nRT/P=[18000/(2×96485.3)]×8.314462618×273.15/100.0=2.11844 L≈2.12 L.

This stipulated ideal amount calculation does not establish real electrode selectivity or operating voltage. A real aqueous-cell model also needs activities, pH, electrodes, overpotentials and competing water oxidation/reduction. Sodium is not simply plated as stable metal from ordinary aqueous NaCl.