Cell Potentials and Good Batteries

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

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

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

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 = +0.34 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 = -0.76 V Element “loses” – negative voltage

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

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

Eocell = Eored + Eoox = + 0.80 – 0.34 + 0.46 What is the voltage for a silver-copper cell? Ag+(aq) + 1e– → Ag(s) E° = +0.80 V red Cu2+(aq) + 2e– → Cu(s) E° = 0.34 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 = + 0.80 – 0.34 + 0.46 spontaneous

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?

+ 2.26 [ ] ×2 Au3+(aq) + 3e– → Au(s) E° = +1.50 V red [ ] ×3 [ ] ×2 Au3+(aq) + 3e– → Au(s) E° = +1.50 V red [ ] ×3   Zn2+(aq) + 2e– → Zn(s) E° = 0.76 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)

2 Ag+(aq) + Cu(s) → 2 Ag(s) + Cu2+(aq) Ag(s) + Cu2+(aq) → no reaction Eoc = +0.46 no reaction Eoc = - 0.46 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

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

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