Given Eo Zn+2/Zn = -0.76Kf [Cu(NH3)4]2+ = 4*10^11Eo Cu2+/cu = 0.34anode - 1 L of 2M Znso4cathode - 1L of 0.2 M Cuso4Que. when 1 mole of nH3 is added to cathode compartment than emf of cell isans. 0.81 please explain how??

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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