Mission: Electrochemistry

Metal + Non-Metal Redox

100

Question: 2Mg + O₂ → 2MgO. Which element is oxidized?

Magnesium

200

Question: 2Na + Cl₂ → 2NaCl. Which element is reduced?

Chlorine

300

Question: Ca + S → CaS. Identify the oxidized and reduced species.

Oxidized = Calcium, Reduced = Sulfur

400

Question: 2Al + 3Br₂ → 2AlBr₃. Write the oxidation half-equation.

Al → Al³⁺ + 3e⁻

500

Question: 2Al + 3Br₂ → 2AlBr₃. Identify the oxidized species, reduced species, write BOTH half-equations, and identify the oxidizing agent.

Oxidized = Al, Reduced = Br₂, Oxidation: Al → Al³⁺ + 3e⁻, Reduction: Br₂ + 2e⁻ → 2Br⁻, Oxidizing Agent = Br₂

Halogen Displacement

100

Question: Cl₂ + 2KBr → 2KCl + Br₂. Which halogen is more reactive?

Chlorine

200

Question: Br₂ + 2KI → 2KBr + I₂. Which element is oxidized?

Iodine (I⁻)

300

Question: Cl₂ + 2KBr → 2KCl + Br₂. Which species is reduced?

Chlorine (Cl₂)

400

Br₂ + 2KI → 2KBr + I₂. Write the oxidation half-equation.

2I⁻ → I₂ + 2e⁻

500

Question: Cl₂ + 2KBr → 2KCl + Br₂. Identify the oxidized species, reduced species, spectator ions, write BOTH half-equations, and write the net ionic equation.

Oxidized = Br⁻, Reduced = Cl₂, Spectator = K⁺, Oxidation: 2Br⁻ → Br₂ + 2e⁻, Reduction: Cl₂ + 2e⁻ → 2Cl⁻, Ionic Equation: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂

Voltaic Cells Basics

100

Question: In a voltaic cell, where does oxidation occur?

Anode

200

Question: In a voltaic cell, where does reduction occur?

Cathode

300

Electrons flow from the _____ to the _____.

Anode to Cathode

400

Question: What type of energy is converted into electrical energy in a voltaic cell?

Chemical energy

500

Question: A voltaic cell consists of a magnesium electrode and a copper electrode. Identify the anode, cathode, direction of electron flow, where oxidation occurs, and where reduction occurs.

Anode = Mg, Cathode = Cu, Electrons flow Mg → Cu, Oxidation at anode, Reduction at cathode

Cell Potential Calculations

100

Question: A Ni/Ag cell is constructed. Which electrode is the cathode?

Ni²⁺ + 2e⁻ → Ni E° = −0.25 V

Ag⁺ + e⁻ → Ag E° = +0.80 V

Silver

200

Question: An Al/Fe cell is constructed. Which electrode is the anode?

Al³⁺ + 3e⁻ → Al E° = −1.66 V

Fe²⁺ + 2e⁻ → Fe E° = −0.44 V

Aluminum

300

Question: A Na/Cu cell is constructed. Calculate E°cell.

• Na⁺ + e⁻ → Na  E° = −2.71 V

• Cu²⁺ + 2e⁻ → Cu E° = +0.34 V

3.05 V

400

Question: A Pb/Fe cell is constructed. Which electrode is the cathode?

• Pb²⁺ + 2e⁻ → Pb E° = −0.13 V

• Fe²⁺ + 2e⁻ → Fe E° = −0.44 V

Lead

500

Question: A voltaic cell is constructed using Mg/Mg²⁺ and Cu/Cu²⁺ half-cells. Given E°(Mg²⁺/Mg) = -2.37 V and E°(Cu²⁺/Cu) = +0.34 V. Identify the anode, cathode, and calculate E°cell. 

Anode = Mg, Cathode = Cu, E°cell = 0.34 − (−2.37) = 2.71 V

Mixed Challenge

100

Question: What does OIL RIG stand for?

Oxidation Is Loss, Reduction Is Gain

200

Question: Zn + Cu²⁺ → Zn²⁺ + Cu. Which species is oxidized?

Zinc

300

Question: A Mg/Cu voltaic cell is built. Which electrode is the anode?

Magnesium

400

Question: Cl₂ + 2Br⁻ → 2Cl⁻ + Br₂. Which species loses electrons?

Br⁻

500

Question: Br₂ + 2I⁻ → 2Br⁻ + I₂. Identify the oxidized species, reduced species, oxidizing agent, reducing agent, write BOTH half-equations, and determine the total number of electrons transferred.

Oxidized = I⁻, Reduced = Br₂, Oxidizing Agent = Br₂, Reducing Agent = I⁻, Oxidation: 2I⁻ → I₂ + 2e⁻, Reduction: Br₂ + 2e⁻ → 2Br⁻, Total electrons transferred = 2