Question

Calculate the standard cell potential for each of the electrochemical cells.

2Ag+(aq)+ Pb(s)---> 2Ag(s) +Pb^2+(aq)
Express your answer using two significant figures.
Ecell=? V


2ClO2(g)+2I-(aq)--->2ClO2-(aq)+I2(s)
Express your answer using two significant figures.
Ecell=?V

O2(g)+4H+(aq)+2Zn(s)--> 2H20(l) + 2Zn2+(aq)
Ecell=?V

Answer 1

(a) Ag+ + e- --> Ag(s) Ecell = .7996 V
Pb2+ + e- --> Pb(s) Ecell= -.13 V
Ecell = cathode - anode = .7996 +.13 = .93 V

(b) ClO2(g) + H+ + e- --> HClO2(aq) Ecell = 1.19 V
I2(s) + 2e- --> 2I- (aq) Ecell = .54 V
Ecell = cathode - anode = 1.19 - .54 = .65 V

(c) O2(g) + 4 H+ + 2e- --> 2H2O (l) Ecell = 1.229 V
Zn2+ + 2e- --> Zn(s) Ecell = -.7618
Ecell = cathode - anode = 1.229 + 0.7618 = 2.0 V (1.9908 V)

Answer 2

a) for the given reaction,
2 Ag⁺ (aq) + Pb (s) ---------> 2 Ag (s) +Pb⁺² (aq)
E_cell value for the given reaction would be -
E_cell= E⁰_oxi+ E⁰_red
From the standard data - Ag⁺/Ag = 0.80 V, Pb⁺²/Pb = -0.13 V
Therefore -
E_cell = 0.80 V + 0.13 V = 0.93 V
c) for the given reaction,
O₂(g) + 4H⁺(aq) + 2 Zn(s) ----> 2H₂O (l) + 2Zn²⁺(aq)
E_cell value for the given reaction would be -
E_cell= E⁰_oxi+ E⁰_red
From the standard data - Zn⁺²/Zn = -0.763 V
Therefore -
E_cell = 1.23 V + 0.763 V = 1.993 V

For b) there are no reduction potentials