Homework Answers
from data table:
Eo(Mg2+/Mg(s)) = -2.37 V
Eo(Ag+/Ag(s)) = 0.80 V
As per given cell notation,
cathode is (Ag+/Ag(s))
anode is (Mg2+/Mg(s))
Eocell = Eocathode - Eoanode
= (0.80) - (-2.37)
= 3.17 V
Mg is converting to Mg2+ and is losing electron
So, electron from Mg to Ag
1 is correct as Mg is anode
2 is wrong as Ag is cathode
Mg is converting to Mg2+ and is losing electron
So, electron from Mg to Ag
3 is correct
4 is correct as calculated above
5 is wrong as oxidation occurs at anode and Mg is anode
Answer: C. 1, 3 & 4 only
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