from data table:
Eo(Cd2+/Cd(s)) = -0.4 V
Eo(Sn2+/Sn(s)) = -0.13 V
As per given reaction/cell notation,
cathode is (Sn2+/Sn(s))
anode is (Cd2+/Cd(s))
Eocell = Eocathode - Eoanode
= (-0.13) - (-0.4)
= 0.27 V
Since Eo cell is positive, the reaction is spontaneous
Answer: yes
17.28. A piece of cadmium is placed in a solution in which [Cd2] - [Sn2 ]...
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