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hjlf
3 years ago
10

Where does oxidation occur in an electrochemical cell?

Chemistry
2 answers:
AlladinOne [14]3 years ago
8 0
Option No. 1 at the cathode in both an electrolytic cell and a voltaic cell

Negative ion arrives at the anode, attracted by electrostatic forces, where they lose electrons. Loss of electrons is defined as oxidation.

The reverse process occur at the negative electrode, where positive ions pick up electrons. They are reduced.
sweet [91]3 years ago
8 0

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

Explanation :

As we know that there are two types of cell which are, electrolytic cell and electrochemical cell or galvanic cell.

Electrolytic cell : It is defined as the 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.

Electrochemical cell : It is defined as the cell in which the chemical reaction occurs with the production of electric potential difference between the two electrodes.  It is also known as voltaic cell.

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

From this we conclude that, the oxidation occurs at the anode in both an electrolytic cell and a voltaic cell.

Hence, the correct option is, (3) at the anode in both an electrolytic cell and a voltaic cell.

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3 years ago
The mass of an atom of element x is equivalent to the total mass of 7 hydrogen atoms.
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7 0
3 years ago
An 11.75 g sample of a common hydrate of cobalt(ii) chloride is heated. after heating, 9.25 g of anhydrous cobalt chloride remai
Irina-Kira [14]
Hydrated salts are when salt crystals have water molecules bound. Anhydrous salts are when the water has been removed.
mass of water removed = hydrated salt - anhydrate salt 
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3 0
3 years ago
For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th
Stolb23 [73]
Given:

Diprotic weak acid H2A:
 
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9. 
Concentration = 0.0650 m 

Balanced chemical equation:

H2A ===> 2H+  + A2- 
0.0650       0        0
-x                2x       x
------------------------------
0.065 - x     2x      x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium. 


5 0
3 years ago
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