Answer:
a. What makes this an oxidation-reduction reaction? (1 point)
Loss of electron by the aluminum and the gain of electron by the silver
b. Write the half-reactions showing the oxidation and reduction reactions. Identify which is the oxidation reaction and which is the reduction reaction. (3 points)
oxidation half reaction: Al → Al3+ + 3e-
Reduction half reaction: Ag+ + e- → Ag
c. What is oxidized in the reaction? What is reduced? (2 points)
Aluminum is oxidized and silver is reduced
d. In this simple electrochemical cell, what functions as the anode? What is the cathode? (3 points)
In a simple electrochemical cell the electrode where oxidation takes place is the anode. And the electrode where reduction reaction happens is the cathode.
e. Is this a galvanic cell or electrolytic cell? Explain your answer. (2 points)
An electrolytic cell because the reaction converts electrical energy into chemical energy
Not sure about e and f!
Heat required = Q = 40 kcal
<h3>Further explanation</h3>
Given
mass of 500 g ice
Required
Heat required
Solution
The heat to change the phase can be formulated :
- Q = m.Lf (melting/freezing)
- Q = m.Lv (vaporization/condensation)
Lf=latent heat of fusion
Lv=latent heat of vaporization
Lf for water = 334 kj/kg=6.01 kJ/mol = 80 cal/g
Phase change(ice to water)
Q= 500 g x 80 cal/g
Q = 40 kcal
Answer:
Explanation:
You need to remember that the oxidation number of H is +1, except when it is in a metal hydrites like NaH, where its oxidation number is -1. Then, the oxidation number of O is -2, but in peroxides is -1. So with these rules you just have to multiply the ox. number with the name of atoms and all the elements in the reaction must sum 0.
