Consider the two reduction half-reactions: Na+(aq) + e− LaTeX: \longrightarrow⟶ Na(s) Eo = −2.71 V Cl2(g) + 2 e− LaTeX: \longrightarrow⟶ 2 Cl−(aq) Eo = 1.36 VUse the electrode potentials above to calculate Eocell and ∆Gorxn for the reaction below, and determine if it is the reaction for a voltaic cell or an electrolytic cell. 2 Na+(aq) + 2 Cl−(aq) LaTeX: \longrightarrow⟶2 Na(s) + Cl2(g)

(i) Δ = -4.07v

(ii)Δ = +785.510 KJ

(iii) The reaction is for an electrolytic cell

Explanation:

In this question, we are tasked with calculating Δ and Δfor the reaction and determine if the reaction is for a voltaic or electrolytic cell.

We proceed as follows;

Firstly, we write the complete  equation of reaction as shown below;

 +      →    +  

We can see that chlorine goes from a negative state to zero state. This is indicative of loss of electron which is defined as oxidation

We can also see that sodium went from a positive state to zero state which is indicative of electron gain which is defined as reduction

Mathematically, the value of Δ for the reaction can be calculated using the formula below;

Δ = Δ   -     Δ =  -2.71v - 1.36v = -4.07V

The value of Δ can be calculated as follows;

Δ = -nFΔ

Where n = 2 and F is 96,500C . We plug these values to get;

Δ = -2 × 96,500C× -4.07V  = 785,510J = 785.510KJ (1KJ = 1000J)

Now, we answer if this reaction is for a voltaic or an electrolytic cell;

It is for an electrolytic cell because electrons flow from cathode to anode as indicated by magnitude of the Δ(positive in value)

8

explanation:

each principal energy level above the first contains one s orbital and three p orbitals. a set of three p orbitals, called the p sublevel, can hold a maximum of six electrons. therefore, the second level can contain a maximum of eight electrons - that is, two in the s orbital and 6 in the three p orbitals.

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