1. Cell Potentials and Good Batteries

2. Electrode potential OUTCOME QUESTION(S): C12-6-09 CELLS AND POTENTIALS
Explain the operation of a voltaic (galvanic) cell at the visual, particulate and symbolic level.
Include: writing half-cell reactions and overall reaction
Calculate standard cell potentials and predict spontaneity of reactions.
Compare and contrast voltaic (galvanic) and electrolytic cells.
Vocabulary & Concepts
Electrode potential

3. Spontaneous reactions occur without added energy
2 Ag+(aq) + Cu(s) →
2 Ag(s) + Cu2+(aq)
exothermic
Ag(s) + Cu2+(aq) →
no reaction
*endothermic
Ag+ ions CAN oxidize Cu metal
Cu2+ could NOT oxidize Ag metal
Spontaneous reactions can make good batteries and release energy

4. Cell potential (Eocell)
created measurement of difference of individual electrode potentials
measured in voltage
Remember voltage measures energy per group of electrons
Electrode potentials (Eo)
voltage calculated by placing a substance in a cell to compete against hydrogen
2H+(aq) + 2e– ↔ H2(g)
Hydrogen was chosen as the “standard” to compare against

5. Element was reduced and is stronger than H
(+) Eo potential:
Element was reduced and is stronger than H
X – reduced Hydrogen – oxidized
H2(g) / H+(aq) // Cu2+(aq) / Cu(s)
Eo = V
Element “wins” – positive voltage
(-) Eo potential:
Element was oxidized and is weaker than H
X – oxidized Hydrogen – reduced
Zn(s) / Zn2+(aq) // H+(aq) / H2(g)
Eo = V
Element “loses” – negative voltage

6. Reduction Potential Chart
NO3¯ + 4 H+ + 3e¯  NO(g) + 2 H2O
+0.96
Hg2+ + 2e¯  Hg(l)
+0.85
Ag+ + e¯  Ag(s)
+0.80
1/2 Hg22+ + e¯  Hg(l)
NO3¯ + 2 H+ + e¯  NO2(g) + H2O
+0.78
Fe3+ + e¯  Fe2+
+0.77
I2(s) + 2e¯  2 I¯
+0.53
Cu+ + e¯  Cu(s)
+0.52
Cu2+ + 2e¯  Cu(s)
+0.34
SO42¯ + 4 H+ + 2e¯  SO2(g)+ 2 H2O
+0.17
Sn4+ + 2e¯ Sn2+
+0.15
S + 2 H+ + 2e¯  H2S(g)
+0.14
2 H+ + 2e¯  H2(g)
0.00
Fe3+ + 3e¯  Fe(s)
–0.04
Pb2+ + 2e¯  Pb(s)
–0.13
Sn2+ + 2e¯  Sn(s)
–0.14
Ni2+ + 2e¯  Ni(s)
–0.25
Co2+ + 2e¯  Co(s)
–0.28
Cd2+ + 2e¯  Cd(s)
–0.40
Se + 2 H+ + 2e¯  H2Se(g)
Fe2+ + 2e¯  Fe(s)
–0.44
Cr2+ + 2e¯  Cr(s)
–0.56
Ag2S + 2e¯  2 Ag(s) + S2¯
–0.69
Cr3+ + 3e¯  Cr(s)
–0.74
(+) means a good “potential” to be reduced – substances likely to take electrons
(-) means a bad “potential” to be reduced – substances likely to give electrons

7. Eocell = Eored + Eoox Calculating Cell Potential (Eocell)
Find electrode potentials of each half-cell
Your table lists electrode reduction half-reactions and potentials.
We are going to make some changes to find oxidation potentials
and calculate the voltage by the SUM of the potentials
Eocell = Eored + Eoox
If the cell potential is…
(+) Eocell – spontaneous reaction
(-) Eocell – non - spontaneous reaction

8. Eocell = Eored + Eoox = + 0.80 – 0.34 + 0.46
What is the voltage for a silver-copper cell?
Ag+(aq) + 1e– → Ag(s)
E° = V
red
Cu2+(aq) + 2e– →
Cu(s)
E° = V ox + –
The substance with the lowest reduction potential will be oxidized  reverse the reaction and switch the sign on the potential
Eocell = Eored + Eoox =
– 0.34
+ 0.46
spontaneous

9. For a cell of zinc and gold metal as electrodes:
a) What is the cathode and what is the anode? b) What is the cell potential (voltage)?
c) What is the net reaction? d) What is the line notation for the cell?

10. + 2.26 [ ] ×2 Au3+(aq) + 3e– → Au(s) E° = +1.50 V red [ ] ×3
[ ] ×2
Au3+(aq) + 3e– → Au(s)
E° = V
red
[ ] ×3  
Zn2+(aq) + 2e– →
Zn(s)
E° = V ox + -
Eocell =
+ 2.26
To write the net reaction you will need to balance the electrons lost and gained
Do not multiple the potentials
2 Au3+(aq) + 3 Zn(s) → 2 Au(s) + 3 Zn2+(aq)
Zn(s) / Zn2+(aq) // Au3+(aq) / Au(s)
(oxidized) (reduced)

11. 2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq) Ag(s) + Cu2+(aq) → no reaction
Eoc = +0.46
no reaction
Eoc =
You should be able to predict spontaneity by doing the math for the cell potential
Hg2+ + 2e¯  Hg(l)
+0.85
Ag+ + e¯  Ag(s)
+0.80
1/2 Hg22+ + e¯  Hg(l)
NO3¯ + 2 H+ + e¯  NO2(g) + H2O
+0.78
Fe3+ + e¯  Fe2+
+0.77
I2(s) + 2e¯  2 I¯
+0.53
Cu+ + e¯  Cu(s)
+0.52
Cu2+ + 2e¯  Cu(s)
+0.34
SO42¯ + 4 H+ + 2e¯  SO2(g)+ 2 H2O
+0.17
Sn4+ + 2e¯ Sn2+
+0.15
S + 2 H+ + 2e¯  H2S(g)
+0.14
2 H+ + 2e¯  H2(g)
0.00
Fe3+ + 3e¯  Fe(s)
–0.04
Pb2+ + 2e¯  Pb(s)
–0.13
Sn2+ + 2e¯  Sn(s)
–0.14

12. Will tin react in hydrochloric acid?
spontaneous
Sn(s) + H+(aq) → ?
Eoc = +0.14
Sn(s) + H+(aq) → Sn2+(aq) + H2(g) 2
The chlorine ion (Cl-) in the acid is too electronegative to react – it is not part of this reaction
Hg2+ + 2e¯  Hg(l)
+0.85
Ag+ + e¯  Ag(s)
+0.80
1/2 Hg22+ + e¯  Hg(l)
NO3¯ + 2 H+ + e¯  NO2(g) + H2O
+0.78
Fe3+ + e¯  Fe2+
+0.77
I2(s) + 2e¯  2 I¯
+0.53
Cu+ + e¯  Cu(s)
+0.52
Cu2+ + 2e¯  Cu(s)
+0.34
SO42¯ + 4 H+ + 2e¯  SO2(g)+ 2 H2O
+0.17
Sn4+ + 2e¯ Sn2+
+0.15
S + 2 H+ + 2e¯  H2S(g)
+0.14
2 H+ + 2e¯  H2(g)
0.00
Fe3+ + 3e¯  Fe(s)
–0.04
Pb2+ + 2e¯  Pb(s)
–0.13
Sn2+ + 2e¯  Sn(s)
–0.14

13. Electrode potential CAN YOU / HAVE YOU? C12-6-09 CELLS AND POTENTIALS
Explain the operation of a voltaic (galvanic) cell at the visual, particulate and symbolic level.
Include: writing half-cell reactions and overall reaction
Calculate standard cell potentials and predict spontaneity of reactions.
Compare and contrast voltaic (galvanic) and electrolytic cells.
Vocabulary & Concepts
Electrode potential