EXPERIMENT 1: Identify the oxidation and reduction half-reactions that occur in Cell 5: Mn(s) | Mn(NO3)2 (aq) || Zn(NO3)2(aq) | Zn(s)
Remember to use proper formatting and notation.

Cathode: Mn → Mn²⁺  +  2e⁻    (Oxidation)

Anode: Zn²⁺  +  2e⁻  →  Zn   (Reduction)

Mn | Mn²⁺ ||  Zn²⁺  | Zn

Explanation:

To identify the half reaction we need to see the oxidation states.

Mn(s) → Ground state → Oxidation state = 0

Mn(NO₃)₂ → Mn²⁺  → The oxidation state has increased.

This is the oxidation reaction. It has released two electrons:

Mn → Mn²⁺  +  2e⁻

Zn(NO₃)₂ →  Zn²⁺

Zn → Ground state → The oxidation state was decreased.

This is the reduction reaction. It has gained two electrons:

Zn²⁺  +  2e⁻  →  Zn

Cathode: Mn → Mn²⁺  +  2e⁻

Anode: Zn²⁺  +  2e⁻  →  Zn

Mn | Mn²⁺ ||  Zn²⁺  | Zn

the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

dynamic equilibrium is the answer.