Redox, Cells & Nuclear Chemistry — Chapter Practice, Grade 11

Track electron transfer, electrical energy, isotope change, and statistical decay.

Problem 1

Find nitrogen oxidation number in NO₃⁻ with O=−2.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

5

x−6=−1, x=+5.

Problem 2

Identify the reducing agent in Zn+Cu²⁺→Zn²⁺+Cu.
Answer and explanation

Zn donates electrons and is oxidized, so it is the reducing agent.

Response criteria

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

Problem 3

Find carbon oxidation number in CO₂ using O=−2.
Hints

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

Answer and explanation

4

x−4=0, x=+4.

Problem 4

How many electrons are lost when 2 Mg atoms become 2 Mg²⁺ ions?
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

4 electrons

Each loses 2, total 4.

Problem 5

In 2Mg+O₂→2MgO, which element is reduced? Show the reasoning, not just the conclusion.
Hints

Oxidation and reduction occur together in the complete reaction.

Answer and explanation

Oxygen, changing from 0 to −2 oxidation state; magnesium is oxidized.

Problem 6

What charge-balance check should a redox equation satisfy?
Hints

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

Answer and explanation

Total charge must match on both sides, along with atom counts and balanced electron loss/gain.

Problem 7

Half-reaction balancing includes atoms, charge and the stated mediumHalf-reaction balancing includes atoms, charge and the stated mediumAcidic-medium reduction exampleMnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂OAtoms: Mn 1; O 4; H 8 on each sideCharge: −1 +8 −5 = +2 on each sideBasic-medium balance requires the corresponding OH⁻ account.
Original selected standard acidic half-reaction bookkeeping. It is a paper analysis, not a procedure for handling permanganate or concentrated acid.
Interpret the supplied visual. How does the charge check show that five electrons are required in the displayed acidic half-reaction?
Answer and explanation

The left side without electrons has charge +7, while Mn²⁺ on the right has +2. Five negative electron charges make both +2.

A half-reaction method first balances non-H/O atoms, then uses water and the medium’s proton/hydroxide account to balance O and H, and electrons to balance charge. In the selected acidic permanganate reduction, five electrons are required. A basic-medium answer should not leave an inappropriate net inventory of free H⁺; converting an acidic balance uses matched OH⁻ additions and water cancellation. The method is formal conservation, not proof that the equation is an elementary microscopic step.

The left side without electrons has charge +7, while Mn²⁺ on the right has +2. Five negative electron charges make both +2.

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

For Mg+2Ag⁺→Mg²⁺+2Ag, identify oxidation, reduction, both agents and the electron amount per mole of Mg reacted.
Answer and explanation

Mg goes 0→+2 and is oxidized/reducing agent. Ag⁺ goes +1→0 and is reduced/oxidizing agent. One mole Mg transfers two moles electrons, consumed by two moles Ag⁺ in the selected complete account.

Mg goes 0→+2 and is oxidized/reducing agent. Ag⁺ goes +1→0 and is reduced/oxidizing agent. One mole Mg transfers two moles electrons, consumed by two moles Ag⁺ in the selected complete account.

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.

Problem 9

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 10

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 11

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 12

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 13

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 14

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 15

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 16

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.

Problem 17

An alpha particle is identical to…
  1. a high-energy photon
  2. a single proton
  3. a single electron
  4. a helium-4 nucleus (2 protons and 2 neutrons)
Answer and explanation

a helium-4 nucleus (2 protons and 2 neutrons)

An alpha particle is a helium-4 nucleus, ⁴₂He — two protons and two neutrons.

Problem 18

A 160 g sample of a radioisotope has a half-life of 6 years. How many grams remain after 12 years?
Answer and explanation

40 g

12 ÷ 6 = 2 half-lives, so the sample halves twice: 160 → 80 → 40 g (or 160 ÷ 2² = 40 g).

Problem 19

During beta-minus decay, the atomic number of the daughter nucleus…
  1. increases by 1
  2. decreases by 1
  3. decreases by 2
  4. stays the same
Answer and explanation

increases by 1

In beta-minus decay, a neutron becomes a proton while an electron and antineutrino are emitted, so Z rises by 1 and A stays the same. Beta-plus decay follows a different change.

Problem 20

Explain the difference between nuclear fission and nuclear fusion, and give one real-world example of each.
Answer and explanation

Fission splits a heavy nucleus into lighter ones — for example, uranium-235 splitting inside a nuclear power reactor. Fusion joins light nuclei into a heavier one — for example, hydrogen fusing into helium in the core of the Sun. Both release large amounts of energy.

Problem 21

A 12 g sample of a radioisotope has a half-life of 2 hours. How many grams remain after 6 hours?
Answer and explanation

1.5 g

6 ÷ 2 = 3 half-lives, so the sample halves three times: 12 → 6 → 3 → 1.5 g (or 12 ÷ 2³ = 1.5 g).

Problem 22

In beta-minus decay, how do A and Z change in the basic nuclear equation? Show the reasoning, not just the conclusion.
Hints

Balance both mass number and charge number.

Answer and explanation

A stays unchanged; Z increases by 1 as a neutron converts to a proton with emitted particles.

Problem 23

Why can radiation risk not be ranked from radiation type alone?
Hints

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

Answer and explanation

Dose, energy, exposure route, duration and shielding matter as well as radiation type.

Problem 24

Nucleon balance and mass–energy balance answer complementary questionsNucleon balance and mass–energy balance answer complementary questions²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + 3¹₀nMass numbers: 236 → 141+92+3 = 236Atomic numbers: 92 → 56+36 = 92Energy comes from the complete consistent mass/binding account.
Original selected balanced U-235 fission channel for bookkeeping. Actual fission has a distribution of product channels and neutron yields, not only this one equation.
Interpret the supplied visual. Does equal total mass number prove the precise rest masses before and after are exactly equal?
Answer and explanation

No. Mass number is nucleon count; different binding energies can produce a rest-mass difference and energy transfer.

Fission splits a heavy nucleus into smaller products and can release neutrons; fusion combines light nuclei. Whether a specified channel releases energy depends on the complete initial/final mass and binding-energy account, not the claim that splitting or joining always releases energy. Mass-number sums count nucleons, while precise mass differences connect with E=Δmc². Use either nuclear or atomic masses consistently so electron masses are not accidentally omitted or double-counted. Reaction accessibility and control require a separate mechanism account.

No. Mass number is nucleon count; different binding energies can produce a rest-mass difference and energy transfer.

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 25

Balance the alpha-decay bookkeeping ²³⁸₉₂U → daughter + ⁴₂He. Give daughter A and Z, then explain why the integers alone do not provide released energy.
Answer and explanation

Daughter A=234 and Z=90, corresponding to thorium. Precise consistent masses or binding energies are needed for an energy calculation; mass-number conservation is not equality of exact rest masses.

Daughter A=234 and Z=90, corresponding to thorium. Precise consistent masses or binding energies are needed for an energy calculation; mass-number conservation is not equality of exact rest masses.

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.

Problem 26

80 g parent undergoes two half-lives. Find parent mass.
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

20 g

80/4=20.

Problem 27

Does all parent disappear after two half-lives?
Answer and explanation

No. One-quarter remains on average in a large population.

Response criteria

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

Problem 28

120 g parent undergoes three half-lives. Find remaining parent.
Hints

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

Answer and explanation

15 g

120/8=15.

Problem 29

What changes in beta-minus decay?
Answer and explanation

A neutron becomes a proton, an electron and an antineutrino are emitted; Z increases 1 and A stays the same.

Response criteria

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

Problem 30

A parent fraction is 1/16 with half-life 2 days. Find elapsed time. Show the reasoning, not just the conclusion.
Hints

A half-life is repeated fractional decrease, not fixed mass loss.

Answer and explanation

Four half-lives, so 8 days.

Problem 31

If initial count is 1000 and decay constant is ln2/4 h⁻¹, find count after 8 h using N=N₀e⁻λt.
Hints

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

Answer and explanation

1000e⁻²ln2=250, equivalent to two half-lives.

Problem 32

Logarithms extend the model beyond whole half-life stepsLogarithms extend the model beyond whole half-life stepsN/N₀ = 0.30; t½ = 10.0 dayst = t½ × ln(N/N₀) / ln(1/2)t ≈ 17.4 daysA fractional number of half-lives is physically meaningful.
Original selected 30% remaining example with a hypothetical ten-day half-life; not an age estimate for a historical object.
Interpret the supplied visual. Why is the time positive when both logarithms in the selected expression are negative?
Answer and explanation

Their ratio is positive; the parent fraction has fallen below one over a positive time interval.

A measured parent fraction need not be a simple power of one half. Rearrange the exponential using logarithms to solve t=t½ ln(N/N₀)/ln(1/2). Both the fraction and logarithm arguments are dimensionless. A fraction between zero and one gives positive time because numerator and denominator are both negative. Dating requires an appropriate initial inventory or calibrated initial ratio and a sufficiently closed accounting system; contamination or material exchange can invalidate a simple age inference.

Their ratio is positive; the parent fraction has fallen below one over a positive time interval.

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 33

A modeled isotope has t½=8.00 days. Comparable initial net signal is 800 counts/min. Later gross signal is 250 and matched background 50 counts/min. Find remaining fraction and modeled elapsed time, naming one assumption.
Answer and explanation

Later net signal=200, so fraction=200/800=0.250. That is two half-lives, giving 16.0 days. Assume stable geometry/efficiency and a net signal proportional to the same parent population with an appropriate closed account.

Later net signal=200, so fraction=200/800=0.250. That is two half-lives, giving 16.0 days. Assume stable geometry/efficiency and a net signal proportional to the same parent population with an appropriate closed account.

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.

Browse all courses