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An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)

Soloha48 [4]

Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c. It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

5 0

3 years ago

Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K.

Elis [28]

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

$\Delta G= \Delta H - T * \Delta S$

<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>

Calculating the $\Delta G= -1267 - 473 K* \Delta S$ :

$\Delta G= -1267 - 473 K* \Delta S$

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: $\Delta S>0[/tex}Back to this expression: [tex]\Delta G= -1267 - 473 K* \Delta S$

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

4 0

3 years ago

When one mole of a solute is present in 500 cm3 of solution, then the concentration of the solution is:

Schach [20]

Answer:

0.5M is the answer.

Explanation:

1M solution is the solution containing 1mole solute dissolved per litre of solution.

Using unitary method,

1000cc gives 1M.

1cc gives 1/1000M.

500 cc gives 500/1000M=0.5M

7 0

3 years ago

Two liquids, a and b are immiscible. liquid a has a density of 0.89 g/ml. liquid b has a density of 0.72 g/ml. what would you ex

Burka [1]

In the given above, we have two densities which are 0.89 g/mL and 0.72 g/mL. We are also given that the liquids are immiscible. After the settlement of the liquids, they will form two layers.

The heavier substance, the one which has a higher density will be at the bottom and the lighter substance, the one which has a lower density will be at the top layer.

5 0

3 years ago

Plz answer this i will mark brainly like and ratr

matrenka [14]

Answer:

answer what ?

Explanation:

4 0

2 years ago