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:
Vsln=13.2mL
Explanation:
MwPb(NO3)2=331.2g/mol
⇒ Vsln = (1L Pb(NO3)2/0.200molPb(NO3)2) * (molPb(NO3)2/331.2gPb(NO3)2) * 0.875gPb(NO3)2
⇒ Vsln = 0.0132L (13.2mL)
Answer:
5.35×10²⁴ atoms of Li
Explanation:
To solve this we apply this rule of three:
We know that 1 mol contains 6.02×10²³ atoms
Therefore, 8.9 moles must contain (8.9 . 6.02×10²³) / 1 = 5.35×10²⁴ atoms
One mol is the amount of substance containing as many elemental entities as atoms, which are contained in 12 g of the C¹² isotope
In conclussion 6.02×10²³ atoms of C¹² isotope weighs 12 g