Name: Kohlrausch's Law and Its Applications
Uploaded: 2019-07-19
Duration: 17 min 59 s
Description: Kohlrausch's law helps us in determining the limiting molar conductivities for any electrolyte, degree of dissociation. Let's learn the applications of Kohlr...

Question

Describe the construction of Daniel cell. Write the cell reaction.

Accepted Answer

The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells.

Oxidation half cell: The metallic zinc strip that dips into an aqueous solution of zinc sulphate taken in a beaker.

Reduction half cell: A copper strip that dips into an aqueous solution of copper sulphate taken in a beaker.

Joining the half cell: The zinc and copper strips are externally connected using a wire through a switch (k) and a load (example: voltmeter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing agar-agar gel mixed with an inert electrolyte such as ${Na}_{2} {SO}_{4}$ etc., The ions of inert electrolyte do not react with other ions present in the half cells, and they are not either oxidised (or) reduced at the electrodes. The solution in the salt bridge Voltmeter cannot get poured out, but through which the ions can move into (or) out of the half cells. When the switch (k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redox reactions which are taking place at the respective electrodes.

Anodic oxidation:

The electrode at which the oxidation occur is called the anode. In Daniel cell, the oxidation take place at zinc electrode, i.e., zinc is oxidised to ${Zn}^{2 +}$ ions and the electrons. The ${Zn}^{2 +}$ ions enters the solution and the electrons enter the zinc metal, then flow through the external wire and then enter the copper strip. Electrons are liberated at zinc electrode and hence it is negative (– ve).

$Zn (s) \to {Zn}^{2 +} (aq) + 2 e^{-}$ (loss of electron-oxidation)

Cathodic reduction:

As discussed earlier, the electrons flow through the circuit from zinc to copper, where the ${Cu}^{2 +}$ ions in the solution accept the electrons, get reduced to copper and the same get deposited on the electrode. Here, the electrons are consumed and hence it is positive (+ve). ${Cu}^{2 +} (aq) + 2 e^{-} \to Cu (s)$ (gain of electron-reduction)

Salt bridge:

The electrolytes present in two half cells are connected using a salt bridge. We have learnt that the anodic oxidation of zinc electrodes results in the increase in concentration of ${Zn}^{2 +}$ in solution. i.e., the solution contains more number of ${Zn}^{2 +}$ ions as compared to ${SO}_{4}^{2 -}$ and hence the solution in the anodic compartment would become positively charged. Similarly, the solution in the cathodic compartment would become negatively charged as the ${Cu}^{2 +}$ ions are reduced to copper i.e., the cathodic solution contain more number of ${SO}_{4}^{2 -}$ ions compared to ${Cu}^{2 +}$ .

Completion of circuit:

Electrons flow from the negatively charged zinc anode into the positively charged copper electrode through the external wire, at the same time, anions move towards anode and cations are move towards the cathode compartment. This completes the circuit.

Consumption of Electrodes:

As the Daniel cell operates, the mass of zinc electrode gradually decreases while the mass of the copper electrode increases and hence the cell will function until the entire metallic zinc electrode is converted in to ${Zn}^{2 +}$ the entire ${Cu}^{2 +}$ ions are converted in to metallic copper. Daniel cell is represented as

Describe the construction of Daniel cell. Write the cell reaction.

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