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Given Eo Zn+2/Zn = -0.76 Kf [Cu(NH3)4]2+ = 4*10^11 Eo Cu2+/cu = 0.34 anode - 1 L of 2M Znso4 cathode - 1L of 0.2 M Cuso4 Que. when 1 mole of nH3 is added to cathode compartment than emf of cell is ans. 0.81 please explain how??

Given Eo Zn+2/Zn = -0.76
Kf [Cu(NH3)4]2+ = 4*10^11
Eo Cu2+/cu = 0.34
anode - 1 L of 2M Znso4
cathode - 1L of 0.2 M Cuso4

Que. when 1 mole of nH3 is added to cathode compartment than emf of cell is
ans. 0.81 please explain how??

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1 Answers

Aman Bansal
592 Points
10 years ago

Dear Ankit,

The question can be done easily by applying the nearst equation and takeing appropriate reaction and correspong E value.

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