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spin [16.1K]
3 years ago
5

At which electrode does oxidation occur in a

Chemistry
2 answers:
Verdich [7]3 years ago
7 0

Answer : The correct option is, (3) the anode in both a voltaic cell and an  electrolytic cell.

Explanation :

As we know that there are two types of cell which are, electrolytic cell and voltaic cell.

Electrolytic cell : It is a cell in which the chemical reaction occurs by passing of the current from external source.

In this electrolytic cell, the oxidation occurs at anode which is a positive electrode and reduction occurs at cathode which is a negative electrode.

Voltaic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the galvanic cell or electrochemical cell.

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

Hence, the oxidation occur in a  voltaic cell and in an electrolytic cell at the anode in both a voltaic cell and an  electrolytic cell.

Trava [24]3 years ago
5 0
The right answer for the question that is being asked and shown above is that: "(2) the cathode in a voltaic cell and the anode in an electrolytic cell." At the status of electrode does oxidation occur in a voltaic cell and in an electrolytic cell is that the cathode in a voltaic cell and the anode in <span>an electrolytic cell</span>
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During the chemical reaction given below 21.71 grams of each reagent were allowed to react. Determine how many grams of the exce
swat32

Answer: 16.32 g of O_2 as excess reagent are left.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

\text{Moles of} SO_2=\frac{21.71g}{64g/mol}=0.34mol

\text{Moles of} O_2=\frac{21.71g}{32g/mol}=0.68mol

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)  

According to stoichiometry :

2 moles of SO_2 require = 1 mole of O_2

Thus 0.34 moles of SO_2 will require=\frac{1}{2}\times 0.34=0.17moles  of O_2

Thus SO_2 is the limiting reagent as it limits the formation of product and O_2 is the excess reagent.

Moles of O_2 left = (0.68-0.17) mol = 0.51 mol

Mass of O_2=moles\times {\text {Molar mass}}=0.51moles\times 32g/mol=16.32g

Thus 16.32 g of O_2 as excess reagent are left.

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2 years ago
MethaneProteinAlcohol can be found in beer, wine and liquor.
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The correct answer is alcohol. It is the common component in beer, wine and any liquor. Usually, alcohol is produced by fermentation of organic products containing glucose to produce alcohol, specifically ethanol, as the important product and the by-products water and carbon dioxide. 
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Hydrogen

Explanation:

Hydrogen can never be central atom despite its low electronegativity

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