Reduction Potential and Cells

Define standard electrode potential Calculate standard cell potentials given standard electrode potentials Predict the spontaneity of reactions using standard electrode potentials Additional KEY Terms

Spontaneous reactions occur without added energy 2 Ag+(aq) + Cu(s) → Ag(s) + Cu2+(aq) → 2 Ag(s) + Cu2+(aq) no reaction Ag+ ions can oxidize Cu metal. Cu2+ cannot oxidize Ag metal.

Cell potential (Eocell) measured in terms of cell voltage Used to be called Emf – electromotive force found using electrode reduction potentials Electrode potentials (Eo) voltage from placing each substance in a cell to compete with hydrogen 2H+(aq) + 2e– ↔ H2(g) Hydrogen was chosen as the “standard” to compare things

X – oxidized Hydrogen – reduced (+) Eo reduced - accept electrons from H+ X – reduced Hydrogen – oxidized H2(g) / H+(aq) // Cu2+(aq) / Cu(s) Eo = +0.34 V (-) Eo oxidized – donated electrons to 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 more likely to give electrons

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

E°cell = E°ox + E°red = - 0.34 + 0.80 + 0.46 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 E°cell = E°ox + E°red = - 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 + - E°cell = + 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 E°c = +0.46 no reaction E°c = - 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

Which is reduced and which is oxidized? Will tin strips in hydrochloric acid react? Sn(s) + H+(aq) → ?? E°c = +0.14 Sn(s) + H+(aq) → Sn2+(aq) + H2(g) 2 Which is reduced and which is oxidized? Will it be spontaneous? 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? Define standard electrode potential Calculate standard cell potentials given standard electrode potentials Predict the spontaneity of reactions using standard electrode potentials Additional KEY Terms