For Zn+Ag2O = ZnO+Ag Write out Nerst equation for both half cells (find E)Predict how the voltage of the battery will change as the starting concentrations decrease. Draw the galvonic cell for the reaction. Find Eo also.
The half reactions will be
Oxidation : Zn --> Zn+2 + 2e [anode]
Reduction : Ag+ + e --> Ag(s) [cathode]
Overall reaction will be
Zn(s) + 2Ag+ --> Zn+2 + 2Ag(s)
Cathodic nernst equation will be
Ecathode = E0cathode - 0.0592 / 2 log 1/[Ag+]^2
Anodic nernst equation will be
Eanode = E0cathode - 0.0592 / 2 log 1/[Zn+2]
Ecell = E0cell - 0.0592 /2 9log[[Zn+2] / [Ag+]^2
So as concentration is decreased the initial concentration of Ag+ will decrease and i will increase the EMF of cell.
E0cell = E0cathode - E0anode
E0cell = 0.80 -(-0.76) = +1.56
The cell can be written as
Zn(s) / ZnO (M1) // Ag2O(M2) / Ag(s)
Nernst equation is
Ecell = E0cell - 0.0592 / n log Q
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