Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
Answer:
Basically, the more hot the state is, the more kinetic energy it will have. This means that answer D. would be right, as it goes from coldest to hottest states!
Answer:
intensity
Explanation:
Intensity has no affect on whether or not the photoelectric effect occurs. The determining property is frequency and since frequency and wavelength are inversely proportional, wavelength matters as well
Which has the highest electronegativity value?
A
hydrogen
B
calcium
C
helium
D
fluorine d because fluorine has a higher group number
C.
centi- is essentially 10^2 of one meter.
If you had 100m, multiplying 100 by 10^2 (or 100) would give you 10000 cm.
