Answer:
E° = 1.24 V
Explanation:
Let's consider the following galvanic cell: Fe(s) | Fe²⁺(aq) || Ag⁺(aq) | Ag(s)
According to this notation, Fe is in the anode (where oxidation occurs) and Ag is in the cathode (where reduction occurs). The corresponding half-reactions are:
Anode: Fe(s) ⇒ Fe²⁺(aq) + 2 e⁻
Cathode: Ag⁺(aq) + 1 e⁻ ⇒ Ag(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an
E° = 0.80 V - (-0.44 V) = 1.24 V
-Positively charged nucleus
-Empty spaced
-Dense core
Explanation:
A covalent bond is a chemical bond that involves the sharing of electron pairs between atoms
*Answer:
Option A: 59.6
Explanation:
Step 1: Data given
Mass of aluminium = 4.00 kg
The applied emf = 5.00 V
watts = volts * amperes
Step 2: Calculate amperes
equivalent mass of aluminum = 27 / 3 = 9
mass of deposit = (equivalent mass x amperes x seconds) / 96500
4000 grams = (9* amperes * seconds) / 96500
amperes * seconds = 42888888.9
1 hour = 3600 seconds
amperes * hours = 42888888.9 / 3600 = 11913.6
amperes = 11913.6 / hours
Step 3: Calculate kilowatts
watts = 5 * 11913.6 / hours
watts = 59568 (per hour)
kilowatts = 59.6 (per hour)
The number of kilowatt-hours of electricity required to produce 4.00kg of aluminum from electrolysis of compounds from bauxite is 59.6 kWh when the applied emf is 5.00V
