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: You started with 8
Explanation: the amount of products is equal to the amount of reactants
All of the above. If you are going to narrow it down, it would be high voltage and radioactivity.
Gay-Lussac's law gives the relationship between pressure and temperature of gas. For a fixed amount of gas, pressure is directly proportional to temperature at constant volume.
P/T = k
where P - pressure , T - temperature and k - constant
parameters for the first instance are on the left side and parameters for the second instance are on the right side of the equation
substituting the values in the equation
T = 4342 K
initial temperature was 4342 K
Valence electrons are the electrons in the outermost shell of an element on the periodic table. Atoms want to be able to have a full outer shell and they can share or trade electrons in order to achieve this. Valence electrons are also super super important in chemical reactions. The number of valence electrons determines what group that specific atom or element is in on the periodic table. This affects the reactivity of the element.