ElectrochemistryHard
Question
The emf of the cell, Ni | Ni2+ (1.0 M) || Ag+ (1.0M) | Ag [Eo for Ni2+ / Ni = - 0.25 volt, Eo 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
Ecell = EoNi / Ni2+ + EoAg+/Ag
= 0.25 + 0.80 = 1.05 Volt.
= 0.25 + 0.80 = 1.05 Volt.
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