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3 years ago

Plz help me with this question

Chemistry

1 answer:

miskamm [114]3 years ago

5 0

Answer:

<h2>Here is your answer. I hope it will help you . </h2>

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

faltersainse [42]

Answer:

For A: The standard cell potential of the reaction is 4.4 V

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

For C: The reaction is spontaneous as written.

Explanation:

For A:

The given chemical reaction follows:

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

The given half reaction follows:

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

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

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

For B:

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 $-8.50\times 10^5J$

For C:

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.

7 0

3 years ago

A student is studying a sample of carbon dioxide gas inside a small syringe. At constant temperature the syringe has a volume of

Soloha48 [4]

The new pressure inside the syringe will be 1.25 atm

At constant temperatures, the volume of a gas is inversely proportional to its pressure.

Thus: P1V1 = P2V2

In this case, P1 = 3.0 atm, V1 = 89.6 mL, V2 = 215 mL

P2 = P1V1/V2

= 3 x 89.6/215

= 1.25 atm

More on gas laws can be found here: brainly.com/question/1190311

4 0

2 years ago

Drag the item from the idea bank to its corresponding match

babymother [125]

Hope this is helpful

5 0

3 years ago

Last one i think.......

MatroZZZ [7]

Answer is c i believe

4 0

3 years ago

Read 2 more answers

A container of oxygen with a fixed volume has a pressure of 13.0 atm at a temperature of 20 °C. What will the pressure of the ox

Norma-Jean [14]

Given:

P1 = 13.0 atm

T1 = 20 °C

T2 = 102 °C

Required:

P2 of oxygen

Solution:

At constant volume, we can apply Gay-Lussac’s law of pressure and temperature relationship

P1/T1=P2/T2

(13.0 atm) / (20 °C) = P2 / (102 °C)

P2 = 66.3 atm

The answer is not in the choices given.

6 0

3 years ago