Answer:
Internally reversible is the answer.
Explanation:
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
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<u>Answer:</u> The amount of heat required to warm given amount of water is 470.9 kJ
<u>Explanation:</u>
To calculate the mass of water, we use the equation:
Density of water = 1 g/mL
Volume of water = 1.50 L = 1500 mL (Conversion factor: 1 L = 1000 mL)
Putting values in above equation, we get:
To calculate the heat absorbed by the water, we use the equation:
where,
q = heat absorbed
m = mass of water = 1500 g
c = heat capacity of water = 4.186 J/g°C
= change in temperature = 
Putting values in above equation, we get:
Hence, the amount of heat required to warm given amount of water is 470.9 kJ
Answer:
105.8 g of Na would be required
Explanation:
Let's think the reaction:
2Na(s) + Cl₂(g) → 2NaCl (s)
1 mol of chlorine reacts with 2 moles of sodium
Then, 2.3 moles of Cl₂ would react with (2.3 .2) / 1 = 4.6 moles
Let's determine the mass of them.
4.6 mol . 23 g/mol = 105.8 g
