Write the cell notation for the voltaic cell that incorporates each of the following redox reactions:(b) Cu²⁺(aq) + SO₂(g) + 2H₂
1 answer:
The cell notation for the given voltaic cell is
SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)
<h3>What is a voltaic cell?</h3>A voltaic cell often called a galvanic cell, is a type of electrochemical cell.
It is a combination of two metal rods called electrodes.
A half cell is formed when both electrodes are submerged in a solution containing a particular mix of ions.
The electrolyte is the liquid in which both electrodes are submerged and acts as a salt bridge connecting the half cells.
The cell undergoes a chemical reaction known as a redox reaction.
One of the electrodes serves as an anode and undergoes oxidation, while the other electrode serves as a cathode and undergoes reduction.
To write the cell notation, write the half-reaction of the cells.
At the anode, oxidation occurs
At cathode
In the cell notation, the anode is written on the left side, and the cathode is written on the right side. A vertical line differentiates the phases and the double vertical line represents the salt bridge.
The cell notation for the given voltaic cell is
SO₂(g) | SO₄²⁻(aq) || Cu²⁺(aq) | Cu (s)
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1.66 M is the concentration of the chemist's working solution.
<h3>What is molarity?</h3>Molarity (M) is the amount of a substance in a certain volume of solution. Molarity is defined as the moles of a solute per litres of a solution. Molarity is also known as the molar concentration of a solution.
In this case, we have a solution of Zn(NO₃)₂.
The chemist wants to prepare a dilute solution of this reactant.
The stock solution of the nitrate has a concentration of 4.93 M, and he wants to prepare 620 mL of a more dilute concentration of the same solution. He adds 210 mL of the stock and completes it with water until it reaches 620 mL.
We want to know the concentration of this diluted solution.
As we are working with the same solution, we can assume that the moles of the stock solution will be conserved in the diluted solution so:
=
(1)
and we also know that:
n = M x
If we replace this expression in (1) we have:
x
=
x
Where 1, would be the stock solution and 2, the solution we want to prepare.
So, we already know the concentration and volume used of the stock solution and the desired volume of the diluted one, therefore, all we have to do is replace the given data in (2) and solve for the concentration which is :
4.93 x 210 = 620 x
= 1.66 M
This is the concentration of the solution prepared.
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<u>Answer:</u> The concentration of at equilibrium is 0.00608 M
<u>Explanation:</u>
As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.
In the second dissociation, the ions will remain in equilibrium.
We are given:
Concentration of sulfuric acid = 0.025 M
Equation for the first dissociation of sulfuric acid:
0.025 0.025 0.025
Equation for the second dissociation of sulfuric acid:
<u>Initial:</u> 0.025 0.025
<u>At eqllm:</u> 0.025-x 0.025+x x
The expression of second equilibrium constant equation follows:
We know that:
Putting values in above equation, we get:
Neglecting the negative value of 'x', because concentration cannot be negative.
So, equilibrium concentration of sulfate ion = x = 0.00608 M
Hence, the concentration of at equilibrium is 0.00608 M