ElectrochemistryHard

Question

The emf of the cell, Ni | Ni²⁺ (1.0 M) || Ag⁺ (1.0M) | Ag [E^o for Ni²⁺ / Ni = - 0.25 volt, E^o for Ag⁺/Ag = 0.80 volt] is given by -

Options

A.-0.25 + 0.80 = 0.55 volt

B.-0.25 - (+0.80) = -1.05 volt

C.0 + 0.80 - (-0.25) = + 1.05 volt

D.-0.80 - (-0.25) = - 0.55 volt

Solution

E_cell = E^oNi / Ni²⁺ + E^oAg⁺/Ag
= 0.25 + 0.80 = 1.05 Volt.

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