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2 years ago

Consider the following voltaic cell:(b) In which half-cell does oxidation occur?

Chemistry

1 answer:

damaskus [11]2 years ago

6 0

At the anode, half-cell oxidation occurs in a voltaic cell.

<h3>Voltaic Cell Principle</h3>

A voltaic cell generates electricity due to the Gibbs free energy of spontaneous redox processes occurring inside the cell, which is the basis for the voltaic cell's operating principle.

Two half-cells plus a salt bridge make up the voltaic cell. An electrolyte-immersed metallic electrode is present on each side of the cell. These two half-cells are wired together to form a connection to a voltmeter.

<h3>Voltaic Cell Parts</h3>

Copper makes comprises the cathode of a photovoltaic cell. This electrode serves as the cell's positive terminal, where reduction takes place.
Anode: Zink metal makes up this electrode. It creates the cell's negative electrode, where oxidation takes place.
Oxidation and reduction are divided into two discrete parts in two half-cells.
Salt Bridge: It contains the electrolytes needed to finish the circuit in the voltaic cell.
The flow of electrons between the electrodes occurs via the external circuit.

Learn more about Voltaic cells here:-

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Rainbow [258]

Answer:

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3 years ago

A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She k

Sergeeva-Olga [200]

<u>Answer:</u> The partial pressure of oxygen gas is 2.76 bar

<u>Explanation:</u>

To calculate the number of moles, we use the equation given by ideal gas which follows:

$PV=nRT$

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant = $0.0831\text{ L bar }mol^{-1}K^{-1}$

n = Total number of moles = ?

Putting values in above equation, we get:

$5.57bar\times 2.19L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{5.57\times 2.19}{0.0831\times 298}=0.493mol$

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ........(1)

where,

$p_A$ = partial pressure of carbon dioxide = 0.318 bar

$p_T$ = total pressure = 5.57 bar

$\chi_A$ = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

$0.318bar=5.57bar\times \chi_{CO_2}\\\\\chi_{CO_2}=\frac{0.381}{5.57}=0.0571$

Mole fraction of a substance is given by:

$\chi_A=\frac{n_A}{n_A+n_B}$

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas, $\chi_{N_2}=\frac{0.221}{0.493}=0.448$

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

$p_{O_2}=5.57bar\times 0.4949\\\\p_{O_2}=2.76bar$

Hence, the partial pressure of oxygen gas is 2.76 bar

6 0

3 years ago

When solving a problem it is important to identify your given and needed units, but it is also important to understand the relat

Angelina_Jolie [31]

Answer:

The question has some details missing. here are the details ; Given the following ;

1. 43.2 g of tablet with 20 cm3 of space

2. 5 cm3 of tablets weighs 10.8 g

3. 5 g of balsa wood with density 0.16 g/cm3

4. 150 g of iron. With density 79g/cm 3

5. 32 cm3 sample of gold with density 19.3 g/cm3

6. 18 ml of cooking oil with density 0.92 g/ml

Explanation:

<u>Appropriate for calculating mass</u>

32 cm3 sample of gold with density 19.3 g/cm3

18 ml of cooking oil with density 0.92 g/ml

<u>Appropriate for calculating volume</u>

5 g of balsa wood with density 0.16 g/cm3

150 g of iron. With density 79g/cm 3

<u>Appropriate for calculating density</u>

43.2 g of tablet with 20 cm3 of space

5 cm3 of tablets weighs 10.8 g

3 0

3 years ago

If 6.02×1023 atoms of element Y have a mass of 28.09 g, what is the identity of Y?

Mazyrski [523]

Silicon is the element having a mass of 28.09 g

<u>Explanation</u>:

Silicon is the element having an atomic mass of 28.09 g / mol. So 28.09 g of silicon contains 6.023 $\times$ 10^23 atoms. One mole of each element can produce one mole of compound.

The Atomic weight of an element can be determined by the number of protons and neutrons present in one atom of that element. So atomic weight expressed in grams always contain the same number of atoms( 6.023 $\times$ 10^23).

Avagadro number is the number of atoms of 1 mole of any gas at standard temperature and pressure. It has been determined that 6.023 $\times$ 10^23 atoms of an element are equal to the average atomic mass of that element.

8 0

3 years ago

I need it please !!!

anyanavicka [17]

Answer:

im pretty sure its 2...

Explanation:

if its wrong im sorry

if its right brainliest pls?

6 0

3 years ago