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tiny-mole [99]
3 years ago
13

Design a voltaic cell using magnesium as one of the electrodes. Magnesium can be represented as either Metal A or Metal B in the

above drawing. Use metal chlorides as the solutions in the two chambers. For example, magnesium chloride, (MgCl2) will be in solution in the chamber with the magnesium electrode. Use NaNO3 in the salt bridge. Select another element for the other electrode. Explain why you selected this element. Include information about the activity of the metal you select and the need for a spontaneous reaction. Metal A: Metal B: In the drawing, 1. Label the oxidation compartment: 2. Label the reduction compartment. 3. Label the direction of the flow of electrons. 4. Label the flow of the magnesium ions. 5. Label the flow of your selected element's ions. 6. What is leaving the salt bridge in the anode compartment? 7. What is leaving the salt bridge in the cathode compartment? 8. Write the oxidation and reduction half-reactions. 9. Calculate the chemical potential of your cell. Show all of your work.

Chemistry
1 answer:
Daniel [21]3 years ago
6 0

Answer:

Here's what I get  

Explanation:

You haven't shown your drawing, so I will assume that Metal A is the anode and Metal B is the cathode.

I will make a galvanic cell using Mg and Zn as the metals.

I selected Zn because it is common and readily available in the lab.

Zn is lower than Mg in the activity series, so Mg should be able to displace Zn from its salts

The standard reduction potentials are:

                                       <u>  E°/V </u>

Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s);  -0.76

Mg²⁺(aq) + 2e⁻ ⇌ Mg(s); -2.38

The Mg half-reaction has the more negative potential, so it will be the oxidation half-reaction.

8 and 9. Oxidation and reduction half-reactions and cell potential

                                                                                 <u>  E°/V</u>

Oxidation:  Mg(s) ⇌ Mg²⁺(aq) + 2e⁻ ;                      +2.38

<u>Reduction</u><u>: Zn²⁺ (aq) + 2e⁻ ⇌ Zn(s);</u>                         <u>-0.76</u>

                   Mg(s) + Zn²⁺ (aq)  ⇌ Mg²⁺(aq) + Zn(s);  +1.62

The cell potential is positive, so the reaction will be spontaneous.

Mg is the anode, so it is Metal A.

Zn is the cathode, so it is Metal B.

1. The Mg|Mg²⁺ half-cell is the oxidation compartment.

2. The Zn²⁺|Zn half-cell is the reduction compartment.

3. The electrons flow from anode to cathode in the external circuit.

4. The Mg²⁺ ions flow from the Mg through the solution to the salt bridge.

5. The Zn²⁺ ions flow from the solution to the Zn.

6. NO₃⁻ ions flow from the salt bridge into the anode compartment to balance the charge of the developing Zn²⁺ ions.

7. Na⁺ ions flow from the salt bridge into the cathode compartment to replace the charge of the depleted Zn²⁺ ions.

 

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Answer:

pH = 12.15

Explanation:

To determine the pH of the HCl and KOH mixture, we need to know that the reaction is  a neutralization type.

HCl  +  KOH  →  H₂O  +  KCl

We need to determine the moles of each compound

M = mmol / V (mL) → 30 mL . 0.10 M = 3 mmoles of HCl

M = mmol / V (mL) → 40 mL . 0.10 M = 4 mmoles of KOH

The base is in excess, so the HCl will completely react and we would produce the same mmoles of KCl

HCl  +  KOH  →  H₂O  +  KCl

3 m       4 m                       -

             1 m                      3 m

As the KCl is a neutral salt, it does not have any effect on the pH, so the pH will be affected, by the strong base.

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3 0
3 years ago
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2) mass of a penny = 2.50 g

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=> 7.35 * 10^22 kg * 1000 g / kg = 7.35 * 10^ 25 g.

4) find how many times the mass of a penny is contained in the mass of the Moon.

You have to divide the mass of the Moon by the mass of a penny

7.35 * 10^ 25 g / 2.50 g = 2.94 * 10^25 pennies

That means that 2.94 * 10^ 25 pennies have the mass of the Moon, which you can check by mulitiplying the mass of one penny times the number ob pennies: 2.50 g * 2.94 * 10^25 = 7.35 * 10^25.

5) Convert the number of pennies into mole unit. That is using Avogadros's number: 6.022 * 10^ 23

7.35 * 10^ 25 penny * 1 mol / (6.022 * 10^ 23 penny) = 1.22* 10^ 2 mole = 122 mol.

Answer: 122 mol
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