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salantis [7]
3 years ago
10

Consider a redox reaction for which E∘ is a negative number. Part A What is the sign of ΔG∘ for the reaction? What is the sign o

f for the reaction? ΔG∘ > 0 ΔG∘ < 0 Request Answer Part B Will the equilibrium constant for the reaction be larger or smaller than 1? Will the equilibrium constant for the reaction be larger or smaller than 1? K > 1 K < 1 Request Answer Part C Can an electrochemical cell based on this reaction accomplish work on its surroundings? Can an electrochemical cell based on this reaction accomplish work on its surroundings? An electrochemical cell based on this reaction not accomplish work on its surroundings An electrochemical cell based on this reaction accomplish work on its surroundings.
Chemistry
1 answer:
Vsevolod [243]3 years ago
6 0

Answer:

a) ΔG∘ > 0

b) K < 1

c) The electrochemical cell based on this reaction cannot accomplish work on its surroundings.

Explanation:

The Gibb's free energy, ΔG∘, is related to the potential of the cell, E∘ through

ΔG∘ = -nFE∘

where n = number of electrons being gained or lost in the electric cell

F = Amount of Faraday's electricity

a) If E∘ < 0, that is, negative,

ΔG∘ = positive (Since n and F cannot be negative)

That is, ΔG∘ > 0

b) Will the equilibrium constant for the reaction be larger or smaller than 1?

The Gibb's free energy is also related to the equilibrium constant through

ΔG∘ = - nRT In K

where n = number of moles = always positive

R = molar gas constant = always positive

T = absolute temperature in Kelvin = almost always positive too.

Recall that ΔG∘ > 0, that is positive too,

Hence,

- nRT In K = ΔG∘

In K = -(ΔG∘/nRT)

Since all of the quantities on the right hand side are positive parameters,

In K = a negative number

Meaning that K is less than 1.

The equilibrium constant is less than 1.

K < 1

c) Can an electrochemical cell based on this reaction accomplish work on its surroundings?

The Gibb's free energy determines spontaneity.

A process with a negative Gibb's free energy is said to be spontaneous and gives off the free energy as the process proceeds.

A process with a positive Gibb's free energy is said to be non-spontaneous. This one cannot give free energy (work) out to the environment.

For this question, the Gibb's free energy is positive, hence, the electrochemical cell based on this reaction cannot accomplish work on its surroundings.

Hope this Helps!!!

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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
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Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

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In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

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