Cell Potentials and Good Batteries

OUTCOME QUESTION(S): C12-6-07: Vocabulary & Concepts Calculate standard cell potentials and predict spontaneity of reactions. 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) measured in voltage sum of individual electrode potentials Remember voltage measures energy per group of electrons Electrode potentials (Eo) voltage measured 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

X – oxidized Hydrogen – reduced (+) 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 (-) 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

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 = Eoox + Eored Cell potential (Eocell) sum of electrode potentials of each half-cell Eocell = Eoox + Eored Your table lists electrode reduction potentials Oxidation potentials are the reverse – we need to switch the sign If the sum is… (+) Eocell – spontaneous reaction (-) Eocell – non - spontaneous reaction

Eocell = Eoox + Eored = + 0.46 - 0.34 + 0.80 What is the cell potential 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 = Eoox + Eored = - 0.34 + 0.80 + 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? 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

CAN YOU / HAVE YOU? C12-6-07: Vocabulary & Concepts Calculate standard cell potentials and predict spontaneity of reactions. Vocabulary & Concepts Electrode potential