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Mnenie [13.5K]
3 years ago
5

Calculate the standard cell potential given the following standard reduction potentials: Al3++3e−→Al;E∘=−1.66 V Cu2++2e−→Cu;E∘=0

.340 V
Chemistry
1 answer:
Maru [420]3 years ago
3 0

Answer:

Explanation:

To calculate the cell potential we use the relation:

Eº cell = Eº oxidation + Eº reduction

Now in order to determine which of the species is going to be oxidized, we have to remember that the more the value of the reduction potential is negative,  the greater its tendency to be oxidized is. In electrochemistry we use the values of the reductions potential in the tables for simplicity  because the only thing we need to do is change the sign of the reduction potential for the oxized species .

So the species that is going to be oxidized is the Aluminium, and therefore:

Eº cell = -( -1.66 V ) + 0.340 V =  5.06 V

Equally valid is to write the equation as:

Eº cell = Eº reduction for the reduced species - Eº reduction for the oxidized species

These two expressions are equivalent, choose the one you fell more comfortable but be careful with the signs.

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Consider the following equilibrium: 2SO^2(g) + O2(9) = 2 SO3^(g)
saul85 [17]

Answer:

At equilibrium, the forward and backward reaction rates are equal.

The forward reaction rate would decrease if \rm O_2 is removed from the mixture. The reason is that collisions between \rm SO_2 molecules and \rm O_2\! molecules would become less frequent.

The reaction would not be at equilibrium for a while after \rm O_2 was taken out of the mixture.

Explanation:

<h3>Equilibrium</h3>

Neither the forward reaction nor the backward reaction would stop when this reversible reaction is at an equilibrium. Rather, the rate of these two reactions would become equal.

Whenever the forward reaction adds one mole of \rm SO_3\, (g) to the system, the backward reaction would have broken down the same amount of \rm SO_3\, (g)\!. So is the case for \rm SO_2\, (g) and \rm O_2\, (g).

Therefore, the concentration of each species would stay the same. There would be no macroscopic change to the mixture when it is at an an equilibrium.

<h3>Collision Theory</h3>

In the collision theory, an elementary reaction between two reactants particles takes place whenever two reactant particles collide with the correct orientation and a sufficient amount of energy.

Assume that \rm SO_2\, (g) and \rm O_2\, (g) molecules are the two particles that collide in the forward reaction. Because the collision has to be sufficiently energetic to yield \rm SO_3\, (g), only a fraction of the reactions will be fruitful.

Assume that \rm O_2\, (g) molecules were taken out while keeping the temperature of the mixture stays unchanged. The likelihood that a collision would be fruitful should stay mostly the same.

Because fewer \!\rm O_2\, (g) molecules would be present in the mixture, there would be fewer collisions (fruitful or not) between \rm SO_2\, (g) and \rm O_2\, (g)\! molecules in unit time. Even if the percentage of fruitful collisions stays the same, there would fewer fruitful collisions in unit time. It would thus appear that the forward reaction has become slower.

<h3>Equilibrium after Change</h3>

The backward reaction rate is likely going to stay the same right after \rm O_2\, (g) was taken out of the mixture without changing the temperature or pressure.

The forward and backward reaction rates used to be the same. However, right after the change, the forward reaction would become slower while the backward reaction would proceed at the same rate. Thus, the forward reaction would become slower than the backward reaction in response to the change.

Therefore, this reaction would not be at equilibrium immediately after the change.

As more and more \rm SO_3\, (g) gets converted to \rm SO_2\, (g) and \rm O_2\, (g), the backward reaction would slow down while the forward reaction would pick up speed. The mixture would once again achieve equilibrium when the two reaction rates become equal again.

5 0
3 years ago
An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average
Deffense [45]

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

7 0
3 years ago
Fatty acid oxidation occurs in the mitochondrial matrix. However, long-chain fatty acyl-CoA molecules cannot cross the inner mem
GaryK [48]

Answer:

Explanation:

see answer below in the attached file.

4 0
3 years ago
On the cooling curve below, between points Dand Ethe substance is changing from
Irina-Kira [14]
Gas to solid ( a ) is correct i think
6 0
3 years ago
Complete the mechanism for the acid-catalyzed hydrolysis of the epoxide in alcohol by adding any missing atoms, bonds, charges,
Sauron [17]

Answer:

See explanation below

Explanation:

You are missing the structure, therefore, I will do an example with one that I found on another place to try to explain.

This acid mechanism always involves carbocations, and positive charges, never negative because we are in acidic mediums.

In the first step, the lone pairs of the oxigen from the epoxide, substract one hydrogen of the reactant.

Second step, the lone pairs of the oxygen from the reactant, do a nucleophylic attack to the carbon of the epoxide. In this case, it will do it to the most substitued carbon.

Then, in the third step by acid base equilibrium, the hydrogen from the reactant that attacked, is substracted from the molecule by a molecule of water (We are in acid medium, therefore, there is traces of water) and the final structure is formed.

Check picture for mechanism:

4 0
3 years ago
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