The iron half cell is
Fe(s) ==> Fe^2+(aq) + 2e
The Ag half cell is
Ag^+(aq) + e ==> Ag(s)
Overall is
Fe(s) + 2Ag^+(aq) ==> Ag(s) + Fe^2+(aq)
I have written (aq) just to show Ag or Fe is in solution; the usual procedure is to place the concn in M there.
The anode is Fe; the cathode is Ag. Ag plates out on the cathode; (Fe goes into solution at the anode)
Part 2 is the reverse of all of this in an electrolytic cell.
Little confusing!! I need help PLZ!!!
Two half cells in a galvanic cell consist of one iron (Fe(s)) electrode in a solution of iron (II) sulphate (FeSO4(aq)) and a silver (Ag(s)) electrode in a silver nitrate solution.
a. Assume the cell is operating as a galvanic cell, state the overall and half-cell reactions.
Describe what will happen to the mass of the cathode while the cell is operating.
b. Repeat part a), assuming that the cell is operating as an electrolytic cell
1 answer
1 answer