Question

A voltaic cell is made using a lead electrode and copper electrode. The electrolytes for each half cell are lead nitrate and copper nitrate. Write the cell notation for this cell and calculate its cell potential (E°cell).

Answer 1

The cell notation is $Pb_{(s)}$  $Pb^{2+} _{(aq)}$ ║$Cu^{+2}_{aq}$  $Cu_{(s)}$ and the cell potential is 0.464

The reaction occurred while losing of hydron is known as oxidation reaction

We can also tell that the reaction occurred while gaining of oxygen atom is known as oxidation reaction.

The reaction occurred while gaining of hydrogen is known as reduction reaction or we can say that the reaction occurred while losing oxygen atom is known as reduction reaction

An electrochemical cell's cell potential is defined as the difference in potential between two half cells. The electrons' capacity to go from one half cell to the other is what causes the potential difference. As a result of the chemical reaction being a redox reaction, electrons can travel across electrodes.

Calculating the Cell potential

E°cell  = E°(reduction) - E°(oxidation)

           = 0.34 - (-0.0124)

           = 0.464

Hence the cell potential is 0.464

Learn more about Cell potential here

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