Question

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)

Answer 1

Answer:

(i) Δ$E^{o}_{cell} }$ = -4.07v

(ii)Δ$G^{o}_{rxn}$ = +785.510 KJ

(iii) The reaction is for an electrolytic cell

Explanation:

In this question, we are tasked with calculating Δ$E^{o}$ and Δ$G^{o}_{rxn}$ for 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;

$2Na^{+}$  +  $2Cl^{-}$    →   $2Na_{(s)}$  +   $Cl_{2(g)}$

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$E^{o}_{cell} }$ for the reaction can be calculated using the formula below;

Δ$E^{o}_{cell} }$ = Δ$E^{o}_{reduction} }$   -     Δ$E^{o}_{oxidation} }$ =  -2.71v - 1.36v = -4.07V

The value of Δ$G^{o}_{rxn}$ can be calculated as follows;

Δ$G^{o}_{rxn}$ = -nFΔ$E^{o}_{cell} }$

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

Δ$G^{o}_{rxn}$ = -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$G^{o}_{rxn}$(positive in value)