Daniel Cell Problems Assignment Help

Electrostatics - Daniel Cell Problems

Daniel Cell

It consists of a copper vessel containing saturated copper sulphate (CuSO4) solution. The copper, a porous pot containing 10% dilute sulphuric acid (called electrolyte) and amalgamated zinc rod (called electrolyte) is placed in the copper vessel and is aptly immersed in a copper sulphate solution. The propos pot prevents the solution from mixing but allows the hydrogen ions to pass through it. A perforated shelf at the top of the vessel, in order to keep the concentration of the copper sulphate solution will be same.

In the propos pot, zinc sulphate is formed. The reactions are as follows:

Zn ---> Zn++ + 2e-

H2SO4 ---> 2H+ + SO--

Zn ++ + SO4 -- ----> Zn SO4

Zn + H2SO4 ---> ZnSO4 + 2H+ + 2e--

The electrons so released are collected by zinc rod making it at negative potential w. r t. electrolyte.

The hydrogen ions (H+) so formed diffuse through the pores of porous pot and interact with copper sulphate forming thereby sulphuric acid and releasing copper ions (Cu++)

2H+ + CuSO4 ---> H2SO4 + Cu++

These positive copper ions extract electrons form copper vessel thereby making it at positive potential to electrolyte convert into copper atoms and are deposited on the copper vessel so that there is no polarization. Thus the copper sulphate acts as the depolarizer. The use of the amalgamated zinc rod avoids the local action in the cell. In this cell as the reaction continues the concentration of copper sulphate solution decreases. Some CuSO4 crystals get dissolved immediately form the perforated shell into CuSO4 solution thus the perforated shelf into CuSO4 solution thus the concentration of CuSO4 is maintained. As the concentration of the copper sulphate solution remains constant when banal cell is in use therefore its remains constant.

The of Daniel cell is = 0.76V – (-0.34V) = 1.11V

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