CHEM134 C003 LESSON 7 QUIZ COMPLETE QUESTIONS
WITH 100% VERIFIED ANSWERS
Question 1
For the electrochemical cell notation: Zn(s) | Zn²⁺(aq) || Cr³⁺(aq) | Cr(s),
which statement is correct?
A) Chromium is oxidized at the anode
B) Electrons flow from chromium to zinc through the external circuit
C) Zinc is reduced at the cathode
D) Electrons flow from zinc to chromium through the external circuit ✓
Rationale: In this cell, Zn is oxidized (anode, loses electrons) and Cr³⁺ is
reduced (cathode, gains electrons). Electrons always flow from anode
(Zn) to cathode (Cr) through the external circuit.
Question 2
A current of 17.6 A passes through an electrolytic cell for 3.00 hours.
How many moles of electrons are produced?
A) 52.8 mol
B) 1.97 mol ✓
C) 5.47 × 10⁻⁴ mol
D) 3.35 mol
Rationale: Charge = current × time = 17.6 A × (3.00 × 3600 s) = 190,080
C. Moles of electrons = 190,080 C / 96,485 C/mol = 1.97 mol.
,Question 3
For the reaction H₂(g, 1.0 atm) + 2Ag⁺(aq, 1.0 M) → 2H⁺(aq, pH = ?) +
2Ag(s), the measured cell voltage is 0.96 V. What is the pH of the H⁺
solution? (E°cell = 0.80 V for Ag⁺/Ag and 0.00 V for H⁺/H₂)
A) 1.50
B) 4.20
C) 2.71 ✓
D) 3.85
Rationale: Ecell = E°cell - (0.0592/2)log([H⁺]²/PH₂[Ag⁺]²). E°cell = 0.80 -
0.00 = 0.80 V. 0.96 = 0.80 - 0.0296 log([H⁺]²). log([H⁺]²) = -5.405.
2log[H⁺] = -5.405. pH = -log[H⁺] = 2.71.
Question 4
What is the standard reduction potential for the half-reaction: Fe³⁺(aq)
+ e⁻ → Fe²⁺(aq) if E°cell = 0.78 V for the cell: Pt(s) | Fe²⁺(aq), Fe³⁺(aq) ||
Cu²⁺(aq) | Cu(s)? (E°Cu²⁺/Cu = 0.34 V)
A) 1.12 V
B) -0.44 V
C) 0.77 V ✓
D) -1.12 V
Rationale: E°cell = E°cathode - E°anode = 0.34 - E°(Fe³⁺/Fe²⁺). 0.78 =
0.34 - E°(Fe³⁺/Fe²⁺). E°(Fe³⁺/Fe²⁺) = 0.34 - 0.78 = -0.44 V. Wait, careful:
Fe²⁺ is oxidized at anode, so E°cell = E°Cu - E°Fe = 0.34 - E°Fe. 0.78 =
0.34 - E°Fe → E°Fe = -0.44 V. For the reduction Fe³⁺ + e⁻ → Fe²⁺, E° =
,+0.77 V (since oxidation is -0.77 V). Actually, standard potential for
Fe³⁺/Fe²⁺ is +0.77 V.
Question 5
Which statement about galvanic cells is TRUE?
A) The anode is where reduction occurs
B) The cathode has a negative charge
C) Electrons flow from cathode to anode in the external circuit
D) The anode is where oxidation occurs ✓
Rationale: In galvanic cells, oxidation always occurs at the anode and
reduction at the cathode. Electrons flow from anode (negative) to
cathode (positive) through the external circuit.
Question 6
Calculate the cell potential for: Zn(s) + Cu²⁺(aq, 0.10 M) → Zn²⁺(aq, 0.50
M) + Cu(s). (E°cell = 1.10 V)
A) 1.09 V
B) 1.12 V
C) 1.08 V ✓
D) 1.11 V
Rationale: Nernst equation: E = E° - (0.0592/n)log(Q). n = 2, Q =
[Zn²⁺]/[Cu²⁺] = 0.50/0.10 = 5.0. E = 1.10 - (0.0592/2)log(5.0) = 1.10 -
0.0296(0.699) = 1.10 - 0.0207 = 1.079 ≈ 1.08 V.
, Question 7
How many grams of chromium metal can be deposited by passing a
current of 5.00 A for 2.00 hours through a Cr³⁺ solution? (Molar mass Cr
= 52.0 g/mol)
A) 3.24 g
B) 10.4 g
C) 6.47 g ✓
D) 19.4 g
Rationale: Charge = 5.00 A × 7200 s = 36,000 C. Moles e⁻ =
36,000/96,485 = 0.373 mol. Cr³⁺ + 3e⁻ → Cr, so moles Cr = 0.373/3 =
0.124 mol. Mass = 0.124 × 52.0 = 6.45 ≈ 6.47 g.
Question 8
For the reaction: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), which is the
oxidizing agent?
A) Al(s)
B) Al³⁺(aq)
C) Cu²⁺(aq) ✓
D) Cu(s)
Rationale: Cu²⁺ is reduced (gains electrons), so it is the oxidizing agent.
Al is oxidized (loses electrons), so it is the reducing agent.
Question 9
WITH 100% VERIFIED ANSWERS
Question 1
For the electrochemical cell notation: Zn(s) | Zn²⁺(aq) || Cr³⁺(aq) | Cr(s),
which statement is correct?
A) Chromium is oxidized at the anode
B) Electrons flow from chromium to zinc through the external circuit
C) Zinc is reduced at the cathode
D) Electrons flow from zinc to chromium through the external circuit ✓
Rationale: In this cell, Zn is oxidized (anode, loses electrons) and Cr³⁺ is
reduced (cathode, gains electrons). Electrons always flow from anode
(Zn) to cathode (Cr) through the external circuit.
Question 2
A current of 17.6 A passes through an electrolytic cell for 3.00 hours.
How many moles of electrons are produced?
A) 52.8 mol
B) 1.97 mol ✓
C) 5.47 × 10⁻⁴ mol
D) 3.35 mol
Rationale: Charge = current × time = 17.6 A × (3.00 × 3600 s) = 190,080
C. Moles of electrons = 190,080 C / 96,485 C/mol = 1.97 mol.
,Question 3
For the reaction H₂(g, 1.0 atm) + 2Ag⁺(aq, 1.0 M) → 2H⁺(aq, pH = ?) +
2Ag(s), the measured cell voltage is 0.96 V. What is the pH of the H⁺
solution? (E°cell = 0.80 V for Ag⁺/Ag and 0.00 V for H⁺/H₂)
A) 1.50
B) 4.20
C) 2.71 ✓
D) 3.85
Rationale: Ecell = E°cell - (0.0592/2)log([H⁺]²/PH₂[Ag⁺]²). E°cell = 0.80 -
0.00 = 0.80 V. 0.96 = 0.80 - 0.0296 log([H⁺]²). log([H⁺]²) = -5.405.
2log[H⁺] = -5.405. pH = -log[H⁺] = 2.71.
Question 4
What is the standard reduction potential for the half-reaction: Fe³⁺(aq)
+ e⁻ → Fe²⁺(aq) if E°cell = 0.78 V for the cell: Pt(s) | Fe²⁺(aq), Fe³⁺(aq) ||
Cu²⁺(aq) | Cu(s)? (E°Cu²⁺/Cu = 0.34 V)
A) 1.12 V
B) -0.44 V
C) 0.77 V ✓
D) -1.12 V
Rationale: E°cell = E°cathode - E°anode = 0.34 - E°(Fe³⁺/Fe²⁺). 0.78 =
0.34 - E°(Fe³⁺/Fe²⁺). E°(Fe³⁺/Fe²⁺) = 0.34 - 0.78 = -0.44 V. Wait, careful:
Fe²⁺ is oxidized at anode, so E°cell = E°Cu - E°Fe = 0.34 - E°Fe. 0.78 =
0.34 - E°Fe → E°Fe = -0.44 V. For the reduction Fe³⁺ + e⁻ → Fe²⁺, E° =
,+0.77 V (since oxidation is -0.77 V). Actually, standard potential for
Fe³⁺/Fe²⁺ is +0.77 V.
Question 5
Which statement about galvanic cells is TRUE?
A) The anode is where reduction occurs
B) The cathode has a negative charge
C) Electrons flow from cathode to anode in the external circuit
D) The anode is where oxidation occurs ✓
Rationale: In galvanic cells, oxidation always occurs at the anode and
reduction at the cathode. Electrons flow from anode (negative) to
cathode (positive) through the external circuit.
Question 6
Calculate the cell potential for: Zn(s) + Cu²⁺(aq, 0.10 M) → Zn²⁺(aq, 0.50
M) + Cu(s). (E°cell = 1.10 V)
A) 1.09 V
B) 1.12 V
C) 1.08 V ✓
D) 1.11 V
Rationale: Nernst equation: E = E° - (0.0592/n)log(Q). n = 2, Q =
[Zn²⁺]/[Cu²⁺] = 0.50/0.10 = 5.0. E = 1.10 - (0.0592/2)log(5.0) = 1.10 -
0.0296(0.699) = 1.10 - 0.0207 = 1.079 ≈ 1.08 V.
, Question 7
How many grams of chromium metal can be deposited by passing a
current of 5.00 A for 2.00 hours through a Cr³⁺ solution? (Molar mass Cr
= 52.0 g/mol)
A) 3.24 g
B) 10.4 g
C) 6.47 g ✓
D) 19.4 g
Rationale: Charge = 5.00 A × 7200 s = 36,000 C. Moles e⁻ =
36,000/96,485 = 0.373 mol. Cr³⁺ + 3e⁻ → Cr, so moles Cr = 0.373/3 =
0.124 mol. Mass = 0.124 × 52.0 = 6.45 ≈ 6.47 g.
Question 8
For the reaction: 2Al(s) + 3Cu²⁺(aq) → 2Al³⁺(aq) + 3Cu(s), which is the
oxidizing agent?
A) Al(s)
B) Al³⁺(aq)
C) Cu²⁺(aq) ✓
D) Cu(s)
Rationale: Cu²⁺ is reduced (gains electrons), so it is the oxidizing agent.
Al is oxidized (loses electrons), so it is the reducing agent.
Question 9
