Electrochemical cells, electrode processes, and cell notation
Question 1
Which statement correctly explains the principle for ‘Translate between a reaction and cell notation’?
(a)Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)Electrons travel through the external circuit from the cathode to the anode during spontaneous operation.
(d)Cell notation conventionally places the reduction half-cell on the left and the oxidation half-cell on the right.
Show answer
(a) Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly.
Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 2
A student is checking Class XII work on ‘Translate between a reaction and cell notation’. Which governing statement should be used?
(a)Cell notation conventionally places the reduction half-cell on the left and the oxidation half-cell on the right.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)Electrons travel through the external circuit from the cathode to the anode during spontaneous operation.
(d)Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly.
Show answer
(d) Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly.
Cell notation places the oxidation half-cell on the left and the reduction half-cell on the right, with phase boundaries and a salt bridge shown correctly. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Electrochemical corrosion and its prevention
Question 3
Which statement correctly explains the principle for ‘Compare protective metal coatings’?
(a)A sacrificial protector should be a more noble metal that is harder to oxidise than iron.
(b)Once scratched, zinc and tin coatings affect the corrosion of exposed iron in exactly the same way.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion.
Show answer
(d) An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion.
An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 4
A student is checking Class XII work on ‘Compare protective metal coatings’. Which governing statement should be used?
(a)Once scratched, zinc and tin coatings affect the corrosion of exposed iron in exactly the same way.
(b)A sacrificial protector should be a more noble metal that is harder to oxidise than iron.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion.
Show answer
(d) An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion.
An intact barrier excludes water and oxygen; if damaged, active zinc remains sacrificial whereas noble tin can accelerate exposed-iron corrosion. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Standard electrode potentials and the electrochemical series
Question 5
Which statement correctly explains the principle for ‘Choose electrodes and predict spontaneity’?
(a)Two tabulated reduction potentials are added directly without reversing the sign of the oxidation half-cell.
(b)The half-cell with the more negative reduction potential is necessarily the spontaneous cathode.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written.
Show answer
(d) The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written.
The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 6
A student is checking Class XII work on ‘Choose electrodes and predict spontaneity’. Which governing statement should be used?
(a)The half-cell with the more negative reduction potential is necessarily the spontaneous cathode.
(b)The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)Two tabulated reduction potentials are added directly without reversing the sign of the oxidation half-cell.
Show answer
(b) The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written.
The more positive reduction potential is the spontaneous cathode and a positive E°cell supports the reaction as written. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Nernst equation, Gibbs energy, and equilibrium
Question 7
At 298 K, what is the emf of Ag(s) | Ag⁺(0.0100 M) || Ag⁺(1.0000 M) | Ag(s)? Treat activities as concentrations relative to 1 M; neglect junction potential.
(a)0 V
(b)0.1182 V
(c)−0.1182 V
(d)0.0591 V
Show answer
(b) 0.1182 V
The standard electrode potentials cancel, but the unequal ion concentrations give a nonzero emf. E = 0 − 0.0591 log(0.0100/1.0000) = 0.1182 V; adding the correction instead gives −0.1182 V.
Question 8
An Ag⁺/Ag concentration cell has ion concentrations 0.0100 M and 1.0000 M at 298 K. How does its emf change when only the concentrated solution is diluted to 0.1000 M? Treat activities as concentrations relative to 1 M; neglect junction potential.
(a)Reverses sign
(b)Decreases
(c)Increases
(d)Remains unchanged
Show answer
(b) Decreases
Dilution reduces the concentration ratio from 100 to 10. The emf decreases from 0.1182 V to 0.0591 V without reversing polarity.
Conductivity, conductance, cell constant, and dilution trends
Question 9
Which statement correctly explains the principle for ‘Use conductivity and cell constant’?
(a)Conductivity equals measured conductance times l/A. The cell constant is found by measuring a known solution.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)Conductivity is obtained by dividing conductance by the cell constant l/A.
(d)Conductance increases in direct proportion to resistance for the same conductivity cell.
Show answer
(a) Conductivity equals measured conductance times l/A. The cell constant is found by measuring a known solution.
Conductivity equals measured conductance times l/A; calibration with a known solution determines the cell constant. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 10
A student is checking Class XII work on ‘Use conductivity and cell constant’. Which governing statement should be used?
(a)Conductivity is obtained by dividing conductance by the cell constant l/A.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)Conductivity equals measured conductance times l/A. The cell constant is found by measuring a known solution.
(d)Conductance increases in direct proportion to resistance for the same conductivity cell.
Show answer
(c) Conductivity equals measured conductance times l/A. The cell constant is found by measuring a known solution.
Conductivity equals measured conductance times l/A; calibration with a known solution determines the cell constant. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Kohlrausch law and ionic-conductivity applications
Question 11
Which statement correctly explains the principle for ‘Find dissociation degree and constant’?
(a)For a weak electrolyte, degree of dissociation is Λm°/Λm and therefore commonly exceeds one.
(b)A weak electrolyte's strongly curved Λm versus √c plot can be linearly extrapolated just like a strong electrolyte.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant.
Show answer
(d) For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant.
For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 12
A student is checking Class XII work on ‘Find dissociation degree and constant’. Which governing statement should be used?
(a)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(b)For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant.
(c)For a weak electrolyte, degree of dissociation is Λm°/Λm and therefore commonly exceeds one.
(d)A weak electrolyte's strongly curved Λm versus √c plot can be linearly extrapolated just like a strong electrolyte.
Show answer
(b) For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant.
For a weak electrolyte α=Λm/Λm°, and concentration with α gives the dissociation constant. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Electrolysis and Faraday laws
Question 13
Which statement correctly explains the principle for ‘Predict aqueous electrolysis products’?
(a)Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present.
(b)In aqueous electrolysis the ions named in the salt must discharge, so water and electrode material need not be considered.
(c)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(d)Reduction occurs at the positive anode of an electrolytic cell because it attracts electrons.
Show answer
(a) Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present.
Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Question 14
A student is checking Class XII work on ‘Predict aqueous electrolysis products’. Which governing statement should be used?
(a)Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)In aqueous electrolysis the ions named in the salt must discharge, so water and electrode material need not be considered.
(d)Reduction occurs at the positive anode of an electrolytic cell because it attracts electrons.
Show answer
(a) Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present.
Compare competing electrode processes, concentrations, overpotentials, and electrode materials rather than selecting any ion present. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Primary cells, secondary batteries, and fuel cells
Question 15
Which statement correctly explains the principle for ‘Explain dry and mercury cells’?
(a)The dry cell uses a zinc container and carbon/MnO2 cathode system; mercury-cell voltage stays nearly constant because dissolved-ion concentrations do not change overall.
(b)The result is fixed by the chemical names alone, so balancing, units, signs, and stated conditions cannot affect it.
(c)Every galvanic battery can be recharged safely by passing current through it in reverse.
(d)A mercury cell keeps nearly constant voltage because its overall reaction steadily changes dissolved-ion concentrations.
Show answer
(a) The dry cell uses a zinc container and carbon/MnO2 cathode system; mercury-cell voltage stays nearly constant because dissolved-ion concentrations do not change overall.
The dry cell uses a zinc container and carbon/MnO2 cathode system; mercury-cell voltage stays nearly constant because dissolved-ion concentrations do not change overall. The other choices each reverse a governing direction, sign, coefficient, unit, or condition.
Every anode is negative because the anode is negative in a galvanic cell.
Why it feels right. The sign in the first cell diagram becomes part of my definition.
Right
Oxidation defines the anode. The sign depends on how the cell operates.
Worked example
Daniell cell
Zn loses electrons→Oxidation at negative anode
Electrolytic cell
Supply pulls electrons from electrode→Oxidation at positive anode
The anode is the oxidation site in both cells. The sign is negative in this galvanic cell and positive in the electrolytic cell.
Zinc oxidation occurs at the negative anode of a Daniell cell.
In an electrolytic cell, the positive terminal of the supply draws electrons from the anode, where oxidation still occurs.
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