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The e.m.f. of the cell
Cu(s) Cu2+aq Ag+ (aq) Ag(s)is 0.46 V. The standard reduction potential of Ag+/Ag is 0.50 V. The standard reduction potential of Cu2+Ι Cu will bea) -0.34 V b) 1.26 Vc) -1.26 V d) 0.34 V

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Solution

Dear Student


The EMF of the Cell, E0 = 0.46 V

The cell reaction is:

Cu (s) + 2Ag+ (aq) ----> Cu2+ (aq) + 2Ag (s)


The half-cell reactions are:

2Ag+ + 2e- ---> 2Ag (s)

Cu (s) ------> Cu2+ (aq) + 2e-


Standard Reduction Potential of Ag+/ Ag = - 0.80 V

[Note: The value of EoAg+/Ag should be +0.80 V, and because they have said reduction potential, then value becomes - 0.80 V]

The question has a printing mistake.

Therefore, Eocell = Eo right - Eo left

= EoAg+/Ag - EoCu2+/Cu

Thus, 0.46 V = - 0.80 V - EoCu2+/Cu

Hence, EoCu2+/Cu = (-0.80 - 0.46) V = -1.26 V




Regards

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