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D Devendra Khairwa
The oxidation equation for the given reaction will be :-                                                           So for oxidation of  1 mol  charge required                                                                                 

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D Devendra Khairwa
According to question the equation of oxidation will be :-                                                   Thus,      for oxidation of   O2- ,  2F charge is required.                                                                                              

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D Devendra Khairwa
The equation for the given question is :-                                          Thus for 1 mol of Al, charge required is 3F. So the required amount of electricity in terms of charge will be :-                                                                             

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D Devendra Khairwa
The equation for the question is given by :-                                                  In this equation, for 1 mol of Ca, 2F charge is required or we can say that for 40 g of Ca charge required is 2F. So,  for 20 g of Ca charge required will be =  F  =   96500 C.

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D Devendra Khairwa
The given reaction can be written as:-                                            Thus charge required in above equation                                                                                                                                 

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D Devendra Khairwa
The nernst equation of the given reaction gives :                     or                            or                            or                            So the required emf of the cell is -1.298 V.

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D Devendra Khairwa
The nernst equation for this reaction gives :-                                       Now for emf, just put all the values.                                      or                                              or                                              Thus emf of the cell is 0.078 V.

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D Devendra Khairwa
The nernst equation for this gives :                                                  This gives :                                         or                                                                    Thus the emf of the given galvanic cell is 0.53 V.

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D Devendra Khairwa
The reaction taking place at both cathode and anode are shown below :-     (i) Cathode reaction :-                                                                    (ii) Anode reaction :-                                                           

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D Devendra Khairwa
The carriers of current in the cell are ions. and Current flows from silver to zinc in the external circuit.

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D Devendra Khairwa
For the given solution :  At cathode :-    Reaction with greater E0 will take place.                                       At anode :-                                      Hence, Cu will get deposited at cathode and Cl2 will be produced from anode.                                     

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D Devendra Khairwa
For the given solution :  At cathode :-    Reaction with greater E0 will take place.                                       At anode :-     Self ionisation of water will take place due to presence of platinum electrode.                                 Hence, H2 gas will be generated at cathode and O2 will be produced from anode.                                     

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D Devendra Khairwa
For the given solution :  At cathode :-    Reaction with greater E0 will take place.                                       At anode :-     Self ionisation will take place due to presence of water.                                 Hence, silver will get deposited at the cathode and O2 will be produced from anode.                                     

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D Devendra Khairwa
For the given solution :  At cathode :-    Reaction with greater E0 will take place.                                       At anode :-                                      Hence, silver will get deposited at the cathode and it will be getting dissolved at anode.                                     

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D Devendra Khairwa
A reaction is feasible only if    is positive. So, anode and cathode reactions will be as follows :-                                                                                                                                  and                                   So this reaction is feasible.

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D Devendra Khairwa
A reaction is feasible only if    is positive. So, anode and cathode reactions will be as follows:-                                                                                                                                  and                                   So this reaction is not feasible.

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D Devendra Khairwa
A reaction is feasible only if    is positive. So, anode and cathode reactions will be as follows :-                                                                                                                                                                  and                                   So this reaction is not feasible.

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D Devendra Khairwa
A reaction is feasible only if    is positive. So, anode and cathode reactions will be as follows :-                                                                                                                                                                  and                                   So this reaction is feasible.

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D Devendra Khairwa
The concept used here will be that a reaction is feasible only if is positive. Anode and cathode reactions will be as follows:-                                                                                                                                                                  So                                   So this reaction is feasible.

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D Devendra Khairwa
Since the cells are connected in series so the current passing through each cell will be equal.(1.5 A) Now we are given that 1.45 g of silver is deposited. So firstly we will consider the cell containing silver.                                                     Since for deposition of 108 g silver 96487 C charge is required, thus for 1.45 g deposition of silver charge required will be:-     ...
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