"cg" is centigram, which is one-hundredth of a gram.
I will first convert from g to cg (multiply by 100), then from mL to L (multiply by 1000).

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
Let's start to understand this question by a simple combustion reaction involving oxidation of Ethane in the presence of Oxygen. When Ethane is burned in the presence of Oxygen it produces Carbon Dioxide and Water respectively. Therefore, the equation is as,
C₂H₆ + O₂ → CO₂ + H₂O
Above reaction shows the reaction and the equation is unbalanced. Balancing chemical equation is important because according to law of conservation of mass, mass can neither be created nor destroyed. Hence, we should balance the number of elements on both side.
LHS RHS
Carbon Atoms 2 1
Hydrogen Atoms 6 2
Oxygen Atoms 2 3
It means this equation is not obeying the law. Now, how to balance? One way is as follow,
C₂H₆ + O₃ → C₂O₂ + H₆O
LHS RHS
Carbon Atoms 2 2
Hydrogen Atoms 6 6
Oxygen Atoms 3 3
We have balanced the equation by changing the subscripts. But, we have messed up the chemical composition of compounds and molecules like Oxygen is converted into Ozone.
Therefore, we will change the coefficients (moles) to balance the equation as,
C₂H₆ + 7/2 O₂ → 2 CO₂ + 3 H₂O
LHS RHS
Carbon Atoms 2 2
Hydrogen Atoms 6 6
Oxygen Atoms 7 7
Now, by changing the coefficients we have balanced the equation without disturbing the chemical composition of compounds and molecules.
Answer:5-3=2
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