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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:
The number of carbon atoms in the container is 1.806 × 10²⁴ or the container contains 1.806 × 10²⁴ atoms of carbon
Explanation:
By Avogadro's number, 1 mole of a substance contains 6.02 × 10²³ particles of the substance
Here we have 0.45 mole of CO₂ contains
0.45 × 6.02 × 10²³ particles of CO₂ that is 2.709 × 10²³ particles of CO₂ or equivalent to 2.709 × 10²³ atoms of Carbon
Similarly, 2.55 moles of CaC₂ contains 2.55 × 6.02 × 10²³ particles of CaC₂ or 1.5351 × 10²⁴ atoms of Carbon
The total number of carbon atoms is therefore;
2.709 × 10²³ + 1.5351 × 10²⁴ = 1.806 × 10²⁴ atoms of carbon.
