What are the half-reactions for a galvanic cell with Zn and Mg electrodes?

the half-reactions

cathode : Zn²⁺ (aq) + 2e⁻ ---> Zn (s)  

anode : Mg (s) → Mg²⁺ (aq) + 2e−

a balanced cell reaction

Zn²⁺(aq) + Mg(s)→ Zn(s) + Mg²⁺ (aq)

Given

Zn and Mg electrodes

Required

The half-reactions for a galvanic cell

Solution

To determine the reaction of a voltaic cell, we must determine the metal that serves as the anode and the metal that serves as the cathode.

To determine this, we can either know from the standard potential value of the cell or use the voltaic series

1. voltaic series

Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au

The more to the left, the metal is more reactive (easily release electrons) and the stronger reducing agent

So the metal on the left will easily undergo oxidation and function as anode

Since Mg is located to the left of Zn, then Mg functions as anode and Zn as a cathode

2. Standard potentials cell of Mg and Zn metals :

Mg2+ + 2e– → Mg E° = -2,35 V

Zn2+ + 2e– → Zn E° = -0,78 V

The anode has a smaller E°, then Mg is the anode and Zn is the cathode.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied