Which is the correct cell notation for the following electrochemical cell? A) Sn(s)|Sn2+(aq,1.0M)||Fe2+(aq,1.0M)|Fe(s) B) Sn(s)|Fe2+(aq,1.0M)||Sn2+(aq,1.0M)|Fe(s) C) Fe(s)|Sn2+(aq,1.0M)||Fe2+(aq,1.0M)|Sn(s) D) Fe(s)|Fe2+(aq,1.0M)||Sn2+(aq,1.0M)|Sn(s) Solution by definition cathode is the one which has more positive reduction potential given Sn+2 (aq) + 2e- ---> Sn (s)   ,,,, Eo = -0.14 V Fe+2 (aq) + 2e- ---> Fe (s) ,,,,,, Eo = -0.44 V so Tin is the cathode here Iron is the anode now oxidation takes place at anode and reduction takes place at cathode anode : Fe(s) ----> Fe+2 (aq) + 2e- cathode : Sn+2 (aq) + 2e- ---> Sn (s) in the cell notation , anode is represented on the left and cathode on the right so Fe(s)|Fe+2(aq)||Sn+2(aq)|Sn(s) so the answer is option D .

1. Which is the correct cell notation for the following electrochemical cell?
A) Sn(s)|Sn2+(aq,1.0M)||Fe2+(aq,1.0M)|Fe(s)
B) Sn(s)|Fe2+(aq,1.0M)||Sn2+(aq,1.0M)|Fe(s)
C) Fe(s)|Sn2+(aq,1.0M)||Fe2+(aq,1.0M)|Sn(s)
D) Fe(s)|Fe2+(aq,1.0M)||Sn2+(aq,1.0M)|Sn(s)
Half Reaction E (V)
Sn2+ (aq)+ 2e- --> Sn(s) -0.14
Fe2+ (aq) + 2e- --> Fe(s) -0.44
Solution
by definition
cathode is the one which has more positive reduction potential
given
Sn+2 (aq) + 2e- ---> Sn (s)Â Â ,,,, Eo = -0.14 V
Fe+2 (aq) + 2e- ---> Fe (s) ,,,,,, Eo = -0.44 V
so
Tin is the cathode here
Iron is the anode
now
oxidation takes place at anode and reduction takes place at cathode

2. anode : Fe(s) ----> Fe+2 (aq) + 2e-
cathode : Sn+2 (aq) + 2e- ---> Sn (s)
in the cell notation , anode is represented on the left and cathode on the right
so
Fe(s)|Fe+2(aq)||Sn+2(aq)|Sn(s)
so
the answer is option D