In an electrolytic cell, where does reduction occur?

Answer: At the cathode (negative electrode) In electrolysis the cathode is connected to the negative terminal of the power source; cations migrate there and gain electrons (reduction).

Answer: Forced non-spontaneous reaction using applied current External voltage drives a reaction with E < 0 (or ΔG 0).

Electrolysis of water with inert electrodes produces

Answer: H₂ at cathode, O₂ at anode Reduction (cathode) → H₂; oxidation (anode) → O₂.

AP Chemistry · Topic 9.10

Electrolysis and Faraday's Law Practice

Q: Electrolysis of water with inert electrodes produces

Answer: H₂ at cathode, O₂ at anode Reduction (cathode) → H₂; oxidation (anode) → O₂.

Part of Applications of Thermodynamics.(ENE-6.D)

Practice questions

Want a predicted score for the whole AP CHEM exam? Take the 20-question diagnostic and Lumi will plan the rest.

Take diagnosticarrow_forward Practice all 9.10

Sample questions

5 of 22 — sign in to practice the rest with adaptive difficulty and mastery tracking.

Sample 1difficulty 2/5
$The schematic shows an electrolytic cell driven by an external DC source.$
$In an electrolytic cell, where does reduction occur?$
- A
  $At the cathode (negative electrode)$
  check_circle
- B
  $In the bulk solution only$
- C
  $At the anode (positive electrode)$
- D
  $In the salt bridge$
Why
$In electrolysis the cathode is connected to the negative terminal of the power source; cations migrate there and gain electrons (reduction).$
View full solution & statsarrow_forward
Sample 2difficulty 2/5
$Electrolysis is$
- A
  $Spontaneous reaction generating current$
- B
  $Forced non-spontaneous reaction using applied current$
  check_circle
- C
  $Distillation$
- D
  $Heating a salt$
Why
$External voltage drives a reaction with E < 0 (or ΔG > 0).$
View full solution & statsarrow_forward
Sample 3difficulty 3/5
$Electrolysis of water with inert electrodes produces$
- A
  $No reaction$
- B
  $H₂ at anode, O₂ at cathode$
- C
  $Cl₂ at anode$
- D
  $H₂ at cathode, O₂ at anode$
  check_circle
Why
$Reduction (cathode) \to H₂; oxidation (anode) \to O₂.$
View full solution & statsarrow_forward
Sample 4difficulty 3/5
$A student electroplates copper from 0.50 M CuSO4 onto a clean copper cathode. A constant current of 0.500 A is passed for 30.0 min. The cathode is dried and reweighed; the mass increases by 0.296 g. F = 96,485 C/mol e-. M(Cu) = 63.55 g/mol.$
$How does the experimental mass of Cu deposited compare to the theoretical mass?$
- A
  $Theoretical = 0.296 g; experimental should be 1/2 because Cu^2+$
- B
  $Theoretical = 0.148 g; efficiency ~200%$
- C
  $Theoretical = 0.296 g; experimental matches within rounding (~100% efficiency)$
  check_circle
- D
  $Theoretical = 0.592 g; efficiency ~50%$
Why
$Q = It = 0.500*1800 = 900 C. mol e- = 900/96485 = 9.328e-3. mol Cu = 9.328e-3/2 = 4.664e-3. mass = 4.664e-3 * 63.55 = 0.296 g.$
View full solution & statsarrow_forward
Sample 5difficulty 3/5
$During operation of a Zn|Zn²⁺ || Cu²⁺|Cu galvanic cell, what happens to electrode masses?$
- A
  $Zn anode mass decreases; Cu cathode mass increases$
  check_circle
- B
  $Zn gains, Cu loses$
- C
  $Both gain mass$
- D
  $Both electrodes lose mass$
Why
$Zn(s) \to Zn²⁺ + 2e⁻ at the anode (mass decreases); Cu²⁺ + 2e⁻ \to Cu(s) at the cathode (mass increases).$
View full solution & statsarrow_forward

Predict my AP scorearrow_forward