The correct answer for the question that is being presented above is this one: "metal." A material you are testing conducts electricity but cannot be pulled into wires. It is most likely a metal. This metal is a good testing for conductivity of an electricity but cannot be pulled into wires.

3 0

3 years ago

Consider the two reduction half-reactions: Na+(aq) + e− LaTeX: \longrightarrow⟶ Na(s) Eo = −2.71 V Cl2(g) + 2 e− LaTeX: \longrig

Juli2301 [7.4K]

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)

7 0

3 years ago

Which of the following molecules is not pyramidal? PH3. CO32– ClO3-. AsH3.

beks73 [17]

It is CO3 2- ,that is the correct answer to your question. :D

8 0

3 years ago