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Question

For the given electrochemical cells :

Cell 1:
EFe2+(aq)/Fe(s)=0.44 V
EH+(aq)/H2(g)=0 V

Cell 2:
EMg2+(aq)/Mg(s)=2.37 V
EH+(aq)/H2(g)=0 V

Choose the correct option:

A
Cell 1 reaction is more spontaneous than cell 2 reaction.
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B
Cell 2 reaction is more spontaneous than cell 1 reaction.
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C
Ecell 1=0.44 V
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D
Both (b) and (c).
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Solution

The correct option is B Cell 2 reaction is more spontaneous than cell 1 reaction.
For cell 1:
Fe gets oxidised to Fe2+ because it has a lower SRP than H2
At anode:
Fe (s)Fe2+(aq)+2e
At cathode:
2H+(aq)+2eH2 (g)

Ecell 1=SRP of cathode - SRP of anode
Ecell 1=EH+(aq)/H2(g)EFe2+(aq)/Fe(s)
Ecell 1=0(0.44)=+0.44 V

Similarly for cell 2:
Mg gets oxidised to Mg2+ because it has a lower SRP than H2
At anode:
Mg (s)Mg2+(aq)+2e
At cathode:
2H+(aq)+2eH2 (g)

Ecell 2=SRP of cathode - SRP of anode
Ecell 2=EH+(aq)/H2(g)EMg2+(aq)/Mg(s)
Ecell 2=0(2.37)=+2.37 V

Higher the Ecell value, more is the spontaneity.
Ecell 2>Ecell 1
Cell 2 reaction is more spontaneous than cell 1 reaction.
Theory :
The magnitude of the cell potential is the measure of the driving force behind a reaction. The larger the value of the cell potential,the farther is the reaction
from equilibrium.





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