Electrochemical Cells & Electrolysis — Practice · Grade 11 Science

Chapter: Redox, Cells & Nuclear Chemistry

What this practice covers

By the end of Electrochemical Cells & Electrolysis you should be able to:

8 questions, numeric and written answer, each with a worked explanation.

Vocabulary used: galvanic cell, electrolysis, Faraday constant.

Problem 1

A 2 A current runs 30 s. Find transferred charge.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

60 C

Q=It=60.

Problem 2

What does a salt bridge carry?
Answer and explanation

Ions move through it to maintain charge balance; external electrons use the wire.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

A 0.50 A current runs 120 s. Find charge transferred.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

60 C

Q=It=.50×120=60.

Problem 4

Why not multiply voltage by a doubled coefficient?
Answer and explanation

Voltage is energy per charge; doubling the reaction doubles both energy and charge.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

Given E°cathode=+0.34 V and E°anode=−0.76 V as reduction potentials, find E°cell. Show the reasoning, not just the conclusion.
Hints

Anode always means oxidation, including electrolytic cells.

Answer and explanation

0.34−(−0.76)=1.10 V.

Problem 6

Why does a salt bridge matter even though external electrons travel through wire?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Ion movement maintains charge balance and completes the internal conducting path; otherwise charge buildup impedes current.

Problem 7

Reduction-potential references combine through a defined sign conventionReduction-potential references combine through a defined sign conventionE°cell = E°reduction,cathode − E°reduction,anodeCu²⁺/Cu: +0.34 V; Zn²⁺/Zn: −0.76 VE°cell = +0.34 − (−0.76) = +1.10 VDo not multiply electrode potential when scaling a half-reaction.
Original standard Zn/Cu reference calculation with rounded potentials and compatible conditions; it is not a guaranteed loaded-device voltage.
Interpret the supplied visual. Why is doubling a two-electron half-reaction not a reason to double its tabulated electrode potential?
Answer and explanation

Potential is energy per charge and remains intensive; scaling changes charge and free-energy amount together, not their ratio.

Standard cell potential E° is found from compatible tabulated reduction potentials: cathode minus anode when both are listed as reductions. The zinc half-reaction is actually reversed during galvanic oxidation, which is handled by this subtraction. Potential is an intensive quantity and is not multiplied by a half-reaction coefficient. The full electron amount does enter ΔG°=−nFE°. Actual voltage depends on activities, temperature and operating losses rather than always matching a standard open-circuit reference value.

Potential is energy per charge and remains intensive; scaling changes charge and free-energy amount together, not their ratio.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

A selected Cu²⁺ deposition receives 0.0100 mol electrons with 100% target current efficiency. Find copper moles and mass using 63.55 g/mol. Explain why a measured lower mass need not violate charge conservation.
Answer and explanation

nCu=0.0100/2=0.00500 mol, giving 0.318 g. Competing reactions, incomplete recovery or measurement error can reduce observed target mass while total transferred charge is still accounted for elsewhere.

nCu=0.0100/2=0.00500 mol, giving 0.318 g. Competing reactions, incomplete recovery or measurement error can reduce observed target mass while total transferred charge is still accounted for elsewhere.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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