Write the net equations, the standard cell potential, and the number of electron
ID: 609457 • Letter: W
Question
Write the net equations, the standard cell potential, and the number of electrons transferred in the following electrochemical cells.
Instructions:
cell Rxn: Ox(1) + Red(2) + other -> Red(1) + Ox(2) + other Eo(V) n Fe/Fe2+//Sn2+/Sn Sn/Sn2+//Ag1+/Ag Pt/I2,I1-//Au3+/AuExplanation / Answer
[Anode reactions] : The electrode where "oxidation" takes place. In your first cell, iron is oxidized from 0 to +2, in the second cell, tin is oxidized from 0 to +2. Anode half reactions for these cells: (1) Fe(s) --------> Fe^2+(aq) + 2e- (2) Sn(s) --------> Sn^2+(aq) + 2e- // : this symbol represents the salt bridge or other means of separating the half cells. [Cathode reactions]: The electrode where "reduction" takes place. In your first cell, tin is reduced from +2 to 0, in the second cell, silver is reduced from +1 to 0. Cathode half reactions for these cells: (1) Sn^2+(aq) + 2e- --------> Sn(s) (2) Ag^1+(aq) + 1e- --------> Ag(s) The overall cell reaction is obtained by adding the anode and the cathode half reactions. Since the number of electrons lost in the cathode should be equal to the electrons gained in the cathode, balance them if necessary 1st Cell: Anode : Fe(s) --------> Fe^2+(aq) + 2e- Cathode : Sn^2+(aq) + 2e- --------> Sn(s) --------------------------------------… Overall : Fe(s) + Sn^2+(aq) ------> Fe^2+(aq) + Sn(s) 2nd Cell: Anode : Sn(s) --------> Sn^2+(aq) + 2e- Cathode : 2Ag^1+(aq) + 2e- --------> 2Ag(s) --------------------------------------… Overall : Sn(s) + Ag^1+(aq) ------> Sn^2+(aq) + 2Ag(s) Standard Reduction Potentials: Fe^2+(aq) + 2e -------> Fe(s) E = - 0.44 V Sn^2+(aq) + 2e- --------> Sn(s) E = - 0.14 V Ag^1+(aq) + 1e- --------> Ag(s) E = + 0.80 V Now the cell potentials can be calculated. Since these are reduction potentials, if you reverse them you get the oxidation potentials. But you must change the sign. However, if you multiply the coefficents of all the terms by a factor, you don't need to multiply the cell voltages, because they are intensive properties. Standard Cell Poentials: 1st Cell: Anode : Fe(s) --------> Fe^2+(aq) + 2e- E = + 0.44 V Cathode : Sn^2+(aq) + 2e- --------> Sn(s) E = - 0.14 --------------------------------------… Overall : Fe(s) + Sn^2+(aq) ------> Fe^2+(aq) + Sn(s) E(cell) = +0.30 V 2nd Cell: Anode : Sn(s) --------> Sn^2+(aq) + 2e- E = + 0.14 V Cathode : 2Ag^1+(aq) + 2e- --------> 2Ag(s) E = + 0.80 V --------------------------------------… Overall : Sn(s) + Ag^1+(aq) ------> Sn^2+(aq) + 2Ag(s) E(cell) = +0.94 V
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