Reaction of galvanic cell
<h3>Further explanation</h3>
Given
Redox reaction between Zn and Mg
Required
Half reaction
Potential of the cell
Solution
Zn²⁺ + 2e -> Zn E ° = -0.76 V
Mg²⁺ + 2e -> Mg E ° = - 2.37 V
The higher E ° of the cell will act as the positive pole/cathode
Electrode Zn as a cathode (reduction) and Mg as an anode (oxidation) (E ° of Zn cells is greater than Mg)
Half reaction
Cathode: Zn²⁺ + 2e ⇒ Zn E ° = -0.76 V
Anode: Mg ⇒ Mg²⁺ + 2e E ° = +2.37 V
------------------------------------------------- ----------- +
Cell reaction: Zn²⁺ + Mg ---> Zn + Mg²⁺ E ° cell = +1.61 V
The reaction occurs spontaneously in the absence of an electric current, thus including galvanic cells/voltaic cells
gdm^-3/gmol^-1
g/g=1 dm^-3/1=dm^3 1/mol^-1=mol^1 1 x dm^-3 x mol^1 mol/dm^3
To convert from g/dm^3 to mol/dm^3 you have to divide g/dm^3 by g/mol or the molecular mass.
The vapor of liquid determines the extent to which molecules in the liquid stay as liquid or escape into the air as gasor vapor
Answer:
127.15 g of ZnCl₂
Solution:
The balance chemical equation is as follow,
<span> Zn + 2 HCl </span>→ ZnCl₂ + H₂
According to equation,
65.38 g (1 mole) of Zn produced = 136.28 g (1 mole) of ZnCl₂
So,
61.0 g of Zn will produce = X g of ZnCl₂
Solving for X,
X = (61.0 g × 136.28 g) ÷ 65.38 g
X = 127.15 g of ZnCl₂