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aniked [119]
3 years ago
8

A voltaic cell is created by using a copper cathode and a magnesium anode. The cathode is immersed in a solution of Cu2 ions, an

d the anode is immersed in a solution of Mg2 ions. A salt bridge of Na2SO4 is also used. What happens to the ions in the salt bridge as the reaction proceeds
Chemistry
1 answer:
Vikki [24]3 years ago
5 0

Answer:

As the reaction proceeds in the given voltaic cell, the Na₂SO₄ present in the salt bridge will dissociate into Na⁺ and SO₄²⁻ ions. As the copper ions in the solution are being deposited on the copper cathode as neutral copper atoms, the solution will become more negative, therefore the Na⁺ ions in the salt bridge will migrate into the the solution in order to maintain electrical neutrality. At the anode, as the Mg metal dissolve into the solution as Mg⁺² ions, the  solution will tend to become more positive. Therefore, the SO₄²⁻ ions present in the salt bridge will migrate into the solution in order to maintain electrical neutrality.

Explanation:

A voltaic or galvanic cell is an example of an electrochemical cell.

An electrochemical cell is a device that produces an electric current from chemical reactions occuring within it.

Electrochemical cells have two electrodes; the anode and the cathode. The anode is defined as the electrode where oxidation occurs while the cathode is the electrode where reduction occurs.

The voltaic cell uses two different metal electrodes each immersed in an electrolyte solution. The two electrodes are connected to each other by means of a wire which allows the flow of electrons from the anode to the cathode. The electrolytes are connected by means of a salt bridge which is a junction that connects the electrolytic solution in the anode and cathode compartment. The salt bridge usually consists of a strong electrolyte like NaCl, KCl, Na₂SO₄, etc.

The electrolyte in the salt bridge serves two purposes: it completes the circuit by providing a path for electron flow and it maintains electrical neutrality in both solutions by allowing ions to migrate between them.

As the reaction proceeds in the given voltaic cell above, the Na₂SO₄ present in the salt bridge will dissociate into Na⁺ and SO₄²⁻ ions. As the copper ions in the solution are being deposited on the copper cathode as neutral copper atoms, the solution will become more negative, therefore the Na⁺ ions in the salt bridge will migrate into the the solution in order to maintain electrical neutrality. Also, at the anode, as the Mg metal dissolve into the solution as Mg⁺² ions, the  solution will tend to become more positive. Therefore, the SO₄²⁻ ions present in the salt bridge will migrate into the solution in order to maintain electrical neutrality.

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Rank the solutions below in order of increasing acidity. (Drag and drop into the appropriate area)
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Answer:

0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl

Explanation:

Strong acids are more acids than weak acids. In the same way, strong bases are more basic than weak bases that are in the same concentration.

Then, the more concentrated acid or base will be more acidic or basic.

CH3COOH. Weak acid

NaOH. Strong base

H2SO4. Strong acid

NH3. Weak base.

HCl. Strong acid

The less acid (More basic):

<h3>0.1 M NaOH, 3 M NH3, 0.01 M CH3COOH, 0.01 M H2SO4, 0.1 M HCl</h3>

Strong base, weak base, weak acid, diluted strong acid, undiluted strong acid

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3 years ago
What is the pH of a solution that has a [OH-] of 5.08x10^-5 M
sattari [20]

Answer:

The pH of a solution that has a [OH-] of 5.08x10^-5 M is 5

Explanation:

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2 years ago
When molten sulfur reacts with chlorine gas, a vile-smelling orange liquid forms that has an empirical formula of SCl. The struc
egoroff_w [7]

Answer:

The structure is shown below.

Explanation:

The formal charge (FC) is the charge that is more close to the actual charge in the real molecules and ions. It can be calculated based on the number of valence electrons (V), the shared electrons (S) and the electrons in the lone pairs (L) by the equation:

FC = V - (L + S/2)

Sulfur is in group 16 of the periodic table, so it has 6 valence electrons, and chlorine is from group 17 of the periodic table, and so it has 7 valence electrons. Chlorine can share only one electron, so it is stable. Sulfur can expand its octet (because it's from the third period) and can have more than 8 electrons when stable.

The possible formulas, from the empiric one, are:

SCl, S₂Cl₂, and S₃Cl₃.

To have FC = 0, chlorine must done only one bond, because S = 2, and L = 6, so:

FC = 7 - (6 + 2/2) = 0

So, it can not be the central atom of a structure. In the SCl, it will hav only a simple bond, so for sulfur, S = 2, and L = 4 (only the lone pairs are counted)

FC = 6 - (4+ 2/2) = +1

For S₂Cl₂, the two sulfurs must be bonded to a simple bond, and each one to one chlorine, thus, for both od them S = 4, and L = 4. so

FC = 6 - (4 + 4/2) = 0

So, it is the correct structure. The lewis structure represents the bonds by lines and the lone pairs of electrons by dots, and it is shown below.

3 0
3 years ago
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