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mamaluj [8]
3 years ago
7

In thermodynamics, we determine the spontaneity of a reaction by the sign of ΔG. In electrochemistry, spontaneity is determined

by the sign of E∘cell. The values of ΔG and E∘cell are related by the following formula: ΔG∘=−nFE∘cell where n is the number of moles of electrons transferred and F=96,500J/V⋅mol e− is the Faraday constant. Part A The standard reduction potentials of lithium metal and chlorine gas are as follows: Reaction, Reduction potential (V) Li+(aq)+e−→Li(s) −3.04
Cl2(g)+2e−→2Cl−(aq) +1.36

In a galvanic cell, the two half-reactions combine to 2Li(s)+Cl2(g)→2Li+(aq)+2Cl−(aq) Calculate the cell potential of this reaction under standard reaction conditions. Express your answer with the appropriate units.

Part B: Calculate the free energy ΔG∘ of the reaction. Express your answer with the appropriate units.

Part C:

What can be said about the spontaneity of this reaction?

The reaction is:

a. spontaneous as written.
b. spontaneous in the reverse direction.
c. at equilibrium.
d. nonspontaneous in either direction.
Chemistry
1 answer:
faltersainse [42]3 years ago
7 0

<u>Answer:</u>

<u>For A:</u> The standard cell potential of the reaction is 4.4 V

<u>For B:</u> The standard Gibbs free energy of the reaction is -8.50\times 10^5J

<u>For C:</u> The reaction is spontaneous as written.

<u>Explanation:</u>

  • <u>For A:</u>

The given chemical reaction follows:

2Li(s)+Cl_2(g)\rightarrow 2Li^+(aq.)+2Cl^-(aq.)

The given half reaction follows:

<u>Oxidation half reaction:</u>  Li(s)\rightarrow Li^+(aq.)+e^-;E^o_{Li^+/Li}=-3.04V ( × 2)

<u>Reduction half reaction:</u>  Cl_2(g)+2e^-\rightarrow 2Cl^-(aq.);E^o_{Cl_2/2Cl^-}=+1.36V

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

Here, chlorine will undergo reduction reaction will get reduced.

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=1.36-(-3.04)=4.4V

Hence, the standard cell potential of the reaction is 4.4 V

  • <u>For B:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

where,

n = number of electrons transferred = 2mol\text{ e}^-

F = Faradays constant = 96500J/V.mol\text{ e}^-

E^o_{cell} = standard cell potential = 4.4 V

Putting values in above equation, we get:

\Delta G^o=-2\times 96500\times 4.4=-849200J=-8.50\times 10^5J

Hence, the standard Gibbs free energy of the reaction is -8.50\times 10^5J

  • <u>For C:</u>

For a reaction to be spontaneous, the standard Gibbs free energy change of the reaction must be negative.

From above, the standard Gibbs free energy change of the reaction is coming out to be negative.

Hence, the reaction is spontaneous as written.

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What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?
AlladinOne [14]

The question is incomplete, here is the complete question:

The given chemical reaction is:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

What is the theoretical yield of Calcium if we begin with 12.6 grams of Aluminium?

<u>Answer:</u> The theoretical yield of copper is 44.48 grams

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

Given mass of aluminium = 12.6 g

Molar mass of aluminium = 27 g/mol

Putting values in equation 1, we get:

\text{Moles of aluminium}=\frac{12.6g}{27g/mol}=0.467mol

The given chemical equation follows:

2Al+3Cu(NO_3)_2\rightarrow 3Cu+2Al(NO_3)_3

By Stoichiometry of the reaction:

2 moles of aluminium produces 3 moles of copper

So, 0.467 moles of aluminium will produce = \frac{3}{2}\times 0.467=0.7005mol of copper

Now, calculating the mass of copper  from equation 1, we get:

Molar mass of copper = 63.5 g/mol

Moles of copper = 0.7005 moles

Putting values in equation 1, we get:

0.7005mol=\frac{\text{Mass of copper}}{63.5g/mol}\\\\\text{Mass of copper}=(0.7005mol\times 63.5g/mol)=44.48g

Hence, the theoretical yield of copper is 44.48 grams

4 0
3 years ago
A 33.0−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h
Lostsunrise [7]

Answer:

THE SPECIFIC HEAT OF THE ALLOY IS 0.9765 J/g K

Explanation:

Mass of alloy = 33 g

Initial temperature of alloy = 93°C

Mass of water = 50 g

Initail temp. of water = 22 °C

Heat capacity of calorimeter = 9.20 J/K

Final temp. = 31.10 °C

specific heat of alloy = unknown

specific heat capacity of water = 4.2 J/g K

Heat = mass * specific heat * change in temperature = m c ΔT

Heat = heat capcity * chage in temperature = Δ H * ΔT

In calorimetry;

Heat lost by the alloy = Heat gained by water + Heat of the calorimeter

                     mc ΔT = mcΔT + Heat capacity * ΔT

33 * C * ( 93 - 31.10) = 50 * 4.2 * ( 31.10 -22) + 9.20 * ( 31.10 -22)

33 * C * 61.9 = 50 * 4.2 * 9.1 + 9.20 * 9.1

2042.7 C = 1911 + 83,72

C = 1911 + 83.72 / 2042.7

C = 1994.72 /2042.7

C =0.9765 J/g K

The specific heat of the alloy is 0.9765 J/ g K

5 0
3 years ago
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