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Law Incorporation [45]
3 years ago
11

Which of the following statements about standard reduction potential and spontaneity are true?

Chemistry
2 answers:
yaroslaw [1]3 years ago
7 0

Answer:

Statements A and C are true.      

Explanation:

<u>Let's evaluate the given statements:</u>

<u />

A) A reaction with negative Gibbs standard free energy (ΔG) will result in a reaction thermodynamically spontaneous, under standard conditions, by the Nernst equation:  

\Delta G = - nFE_{cell}        (1)

<em>where n: is the number of electrons transferred in the cell reaction, F: is the Faraday constant and E: is the cell potential </em>

For an electrochemical reaction to be spontaneous, that is to say, in a galvanic cell, the change in Gibbs free energy must be negative. From equation (1), we have that a negative ΔG will result in a spontaneous electrochemical reaction.  

Hence, statement A is correct, and D is incorrect.

C) A coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous:  

On equation (1), for the Gibbs free energy to be negative and, therefore, the reaction to be thermodynamically spontaneous, the standard reduction potential must be positive.    

Hence, statement C is correct, and B is incorrect.    

I hope it helps you!  

Dafna1 [17]3 years ago
5 0

Answer:

A and C are true , B and D are false

Explanation:

For A)

from the first law of thermodynamics (in differential form)

dU= δQ - δW = δQ - PdV

from the second law

dS ≥ δQ/T

then

dU ≤ T*dS - p*dV

dU - T*dS + p*dV ≤ 0

from the definition of Gibbs free energy

G=H - TS = U+ PV - TS  → dG= dU + p*dV + V*dp - T*dS - S*dT

dG -  V*dp +  S*dT = dU - T*dS + p*dV ≤ 0

dG ≤ V*dp -  S*dT

in equilibrium, pressure and temperature remains constant ( dp=0 and dT=0). Thus

dG ≤ 0

ΔG ≤ 0

therefore the gibbs free energy should decrease in an spontaneous process → A reaction with a negative Gibbs standard free energy is thermodynamically spontaneous under standard conditions

For B) Since the standard reduction potential is related with the Gibbs standard free energy through:

ΔG⁰=-n*F*E⁰

then, when ΔG⁰ is negative , E⁰ is positive and therefore a coupled redox reaction with a positive standard reduction potential is thermodynamically spontaneous.

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The absorbance of a 0.45 mol dm–3 solution of an aromatic amino acid, 3.0 cm thick is 0.22 at a wavelength of 295 nm:
34kurt

Explanation:

a) Using Beer-Lambert's law :

Formula used :

A=\epsilon \times C\times l

where,

A = absorbance of solution = 0.22

C = concentration of solution = 0.45 mol/dm^3=0.45 mol/L=0.45 M

1 dm^3 = 1 L

l = length of the cell = 3.0 cm

\epsilon = molar absorptivity of this solution = ?

Now put all the given values in the above formula, we get the molar absorptivity of this solution.

0.22=\epsilon \times (0.45 M)\times (3.0 cm)

\epsilon=0.163 M^{-1}cm^{-1}

Therefore, the molar absorptivity of this solution is, 1.93\times 10^{4}M^{-1}cm^{-1}

b) A=\log \frac{I_o}{I_t}

T=\frac{I_t}{I_o}

A=\log \frac{1}{T}

A = 2 × 0.22 =0.44

I_o,I_t = Intensities of Incident light and transmitted light respectively

T = Transmittance

0.44=\log \frac{1}{T}

T = 0.3630

c) I_o=x

I_t=65\% of x=0.65 x

Thickness of cell = l' =?

c = 0.75 mol/ dm^3=0.75 mol/L=0.75 M

A=\log \frac{I_o}{I_t}=\epsilon \times C\times l

\log \frac{x}{0.65x}=0.163 M^{-1}cm^{-1}\times 0.45 M\times l'

l' = 1.53 cm

d) No, we cannot calculate the absorbance at 590 nm from the given data. This is because absorbance at this wavelength  can be observe experimentally.

8 0
3 years ago
A student chromatographs a mixture and after developing the spots with a suitable reagent he observes the following
Anika [276]

Rf value is the ratio of the distance traveled by the solute to that of the solvent front on the paper used in chromatographic separation.

From the image it is clear the distance traveled by solvent front = 7.3 cm

Distance traveled by the component -1 of the mixture = 1.4 cm

Distance traveled by the component -2 of the mixture = 3.0 cm

Distance traveled by the component -3 of the mixture = 4.5 cm

Distance traveled by the component -4 of the mixture = 6.5 cm

Rf value of component-1 = \frac{1.4 cm}{7.3 cm} =0.192

Rf value of component-2 = \frac{3.0 cm}{7.3 cm} =0.410

Rf value of component-3 = \frac{4.5 cm}{7.3 cm} =0.616

Rf value of component-4 = \frac{6.5 cm}{7.3 cm} =0.890

b) Samples can be separated from a mixture using chromatography as the relative affinities for the compounds towards the paper (stationary phase) and the solvent(mobile phase) are different. Each component spends different amounts of time on the stationary phase depending on it chemical nature. So, the components in a mixture can be separated based on their polarities and relative degrees of adsorption on the stationary phase.








6 0
3 years ago
Two cylinders are made of the same material. Cylinder A is one-fourth (1/4) the length of cylinder B and it has a radius that is
tankabanditka [31]

Answer:

M_{A} : M_{B}  = 4 : 1

Explanation:

Given that:

Volume of a cylinder = \pir^{2}h

For cylinder A, l_{A} = h_{A} = \frac{1}{4} l_{B} and r_{A} = 4r_{B}.

Volume of cylinder A = \pi(4r_{B}) ^{2} × \frac{1}{4} l_{B}

                                   = 4\pir_{B} ^{2} l_{B}

Volume of cylinder B =  \pi(r_{B}) ^{2} l_{B}

                                  =  \pir_{B} ^{2}l_{B}

To determine the ratio of their masses, density (ρ) is defined as the ratio of the mass (M) of a substance to its volume (V).

i.e    ρ = \frac{M}{V}

Thus, since the cylinders are made from the same material, they have the same density  (ρ). So that;

density of A = density of B

density of A = \frac{M_{A} }{4\pi r_{B} ^{2}l_{B}    }

density of B = \frac{M_{B} }{\pi r_{B} ^{2}l_{B}  }

⇒            \frac{M_{A} }{4\pi r_{B} ^{2}l_{B}    }  =  \frac{M_{B} }{\pi r_{B} ^{2}l_{B}  }

The ratio of mass of cylinder A to that of B is given as;

                       \frac{M_{A} }{M_{B} }  =  \frac{4\pi r_{B} ^{2} l_{B}  }{\pi r_{B} ^{2} l_{B}  }

⇒                        \frac{M_{A} }{M_{B} } = \frac{4}{1}

Therefore, M_{A} : M_{B}  = 4 : 1

8 0
3 years ago
Read 2 more answers
A. 250 Kelvin to Celsius
Liono4ka [1.6K]
250 Kelvin = 76.85 Celsius
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4 years ago
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If 38.7 grams of lithium chlorate decomposes, how many grams of oxygen gas can be produced?
solniwko [45]
LiClO3 has mole of mass of 90 g/mol

So 38.7 g means 0.43 mol

Since in this reaction, LiClO3 : O2 = 2:3, then it will produce 0.43*3/2 = 0.645 mol O2

So it is equivalent to 0.645 * 32 = 20.64 g O2
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3 years ago
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