Answers - Electrochemistry

Arkansas State University
Department of Chemistry
and Physics

Answers

Electrochemistry

1. Write the cell diagram for the Cu⁺²/Cu and Al⁺³/Al half-cells and calculate the E^o_cell for the Cu⁺²/Cu and Al⁺³/Al half-cells.

Write half-reactions with more negative on top:

Al⁺³(aq) + 3 e^- ------->   Al (s) E^o= -1.66 V

Cu⁺²(aq) + 2 e^- -------> Cu (s) E^o= +0.34 V

Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of E^o

Al (s)   ------->    Al⁺³(aq) + 3 e^- E^o= +1.66 V       (anode)

Cu⁺²(aq) + 2 e^- -------> Cu (s) E^o= +0.34 V       (cathode)

Add cell potentials of half-reactions to determine E^o_cell E^o_cell= +2.00 V

cell diagram (anode first):

Al (s)I Al⁺³(aq, 1M)I KCl (sat'd) I Cu⁺²(aq, 1 M) I Cu (s)

2. Determine the E for the Ag⁺/Ag and Sn⁺²/Sn half-cells if the [Ag⁺] = 1.0 M and the [Sn⁺²] = 0.25 M.

Write half-reactions with more negative on top:

Sn⁺²(aq) + 2 e^-   ------->   Sn (s) E^o= -0.14 V

Ag⁺(aq) + 1 e^- -------> Ag (s) E^o= +0.80 V

Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of E^o

Sn (s)   ------->    Sn⁺²(aq) + 2 e^- E^o= +0.14 V       (anode)

Ag⁺(aq) + 1 e^- -------> Ag (s) E^o= +0.80 V      (cathode)

Add cell potentials of half-reactions to determine E^o_cell E^o_cell= +0.94 V

To determine E, use Nernst equation: E = E^o-(0.0591/n)log Q

E = E^o-(0.0591/n)log Q; n = 2 (make e^- lost = e^- gained in the half-reactions);
Q = [Sn⁺²]/ [Ag⁺]²([Sn⁺²] is on the product side, [Ag⁺] is on the reactant side after switching half-reactions)

E = 0.94V -(0.0591 V/2)log {0.25M Sn⁺²/ (1.00 Ag⁺)²} E = 0.958 V

3. Determine the DG^o for the Ag⁺/Ag and Sn⁺²/Sn half-cells.

DG^o= - nFE^o= - 2 mol( 96500 J/V mol)(0.94 V) = -181,420 J = -181.42 kJ

Write half-reactions with more negative on top:
Al⁺³(aq) + 3 e^- -------> Al (s)	E^o= -1.66 V
Cu⁺²(aq) + 2 e^- -------> Cu (s)	E^o= +0.34 V
Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of E^o
Al (s) -------> Al⁺³(aq) + 3 e^-	E^o= +1.66 V (anode)
Cu⁺²(aq) + 2 e^- -------> Cu (s)	E^o= +0.34 V (cathode)

Add cell potentials of half-reactions to determine E^o_cell	E^o_cell= +2.00 V

Write half-reactions with more negative on top:
Sn⁺²(aq) + 2 e^- -------> Sn (s)	E^o= -0.14 V
Ag⁺(aq) + 1 e^- -------> Ag (s)	E^o= +0.80 V
Species on bottom left reacts with species above and to the right, therefore switch top reaction and change sign of E^o
Sn (s) -------> Sn⁺²(aq) + 2 e^-	E^o= +0.14 V (anode)
Ag⁺(aq) + 1 e^- -------> Ag (s)	E^o= +0.80 V (cathode)

Add cell potentials of half-reactions to determine E^o_cell	E^o_cell= +0.94 V

To determine E, use Nernst equation: E = E^o-(0.0591/n)log Q
E = E^o-(0.0591/n)log Q; n = 2 (make e^- lost = e^- gained in the half-reactions); Q = [Sn⁺²]/ [Ag⁺]²([Sn⁺²] is on the product side, [Ag⁺] is on the reactant side after switching half-reactions)
E = 0.94V -(0.0591 V/2)log {0.25M Sn⁺²/ (1.00 Ag⁺)²}	E = 0.958 V