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Describe the construction of Daniel cell. Write the cell reaction. |
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Answer» Solution :Construction of Daniel cell : The separation of half reaction is the basis for the construction of Daniel cell. It consists of two half cells. (i) Oxidation half cell : A metallic zinc strip that dips into an aqueoussolution of zinc sulphate taken in a beaker. (ii) Reduction half cell : A copper strip that dipsinto an aqueous solution of copper sulphate taken in a beaker. (iii) Joining the half cells :The zincand copper strips are externally connected using a wire through a switch (k) and a load (example: volt meter). The electrolytic solution present in the cathodic and anodic compartment are connected using an inverted U tube containing a agar-agar gel mixed with an inert electrolyte such as KCl, `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. Thesolution in the salt bridge cannot get POURED out, butthrough which the ions can MOVE into (or) out of the half cells. (iv) When the switch(k) closes the circuit, the electrons flows from zinc strip to copper strip. This is due to the following redoxreactions which are taking place at the respective electrodes. Cell reaction : Anodicoxidation : The electrode at which the oxidation occuris 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 zincelctrode and hence it is negative (-ve). `Zn_((s)) rarr Zn^(2+)""_((aq))+2e^(-)` `""` (loss of electron-oxidation) (i) Cathodic reduction : The electrons flow through the circuit from zinc to copper, where the `Cu^(2+)` ions in the solution acccept 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))+2e^(-) rarr Cu_((s))` `""` (gain of electron-reduction) (ii) Salt bridge : The electrolytes present in two half cells are connected using a salt bridge. The anodic oxidation of zinc electrodes results in the increase in concentration of `Zn^(2+)` n 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+)`. To maintain the electrical neutrality in both the compartments, the non reactive anions `Cl^(-)` (from KCl taken in the salt bridge) move from the salt bridge and enter into the anodic compartment, at the same time some of the `K^(+)` ions move from the salt bridge into the cathodic compartment. (iii) 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. (iv) 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+)` or the entire `Cu^(2+)` ions are converted in to metallic copper. Unlike Daniel cell, in certain cases, the reactants (or) products cannot serve as electrodes and in such cases inert electrode such as graphite (or) platinum is used which conducts current in the external circuit. 
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