Redox Reactions in Batteries Chem 253 November 11, 2013

Lead-Acid 1859 Gaston Plante’ Discharge – Galvanic Cell (Spontaneous) Negative plate: Pb (0) (s) + HSO 4 - (aq)  Pb (II) SO 4 (s) + H + (aq) + 2e - E 0 = V Positive Plate Pb (IV) O 2 (s) + HSO H + + 2e -  Pb (II) SO 4 (s) + 2H 2 O E 0 = 1.69 V

Cell Reaction Conc. H 2 SO 4 (aq) (6 M) Pb (0) (s) + Pb (IV) O 2 (s) + 2HSO 4 - (aq) + 3H +  2Pb (II) SO 4 (s) + H + (aq) + 2H 2 O E cell = E cath – E anod = 1.69 V – (-0.36) V = 2.05 V Complete discharge H 2 SO 4 (aq) (3 M)

Pb (0) (s) + HSO 4 - (aq)  Pb (II) SO 4 (s) + H + (aq) + 2e - E (V) More Negative More Positive E 0 = V E 0 = 1.69 V Pb (IV) O 2 (s) + HSO H + + 2e -  Pb (II) SO 4 (s) + 2H 2 O Charge Discharge

Pb ElectrodePbSO 4 (s) PbO 2 (s)PbSO 4 (s) Voltmeter or Electrical Load Pb (0) + HSO 4 -  Pb (II) SO 4 + H + + 2e - 6 M H 2 SO 4 Pb (IV) O 2 + HSO H + + 2e -  Pb (II) SO 4 + 2H 2 O HSO 4 - Rxn Zone

Car Battery 6 cell (series) 6 * 2.05 V = 12 V battery

Lead Acid Rate of reaction – requires porous PbSO 4 layers Rxn Rate  electrical current Lead is abundant 99% recycled – none of Li Ion is recyclable 30-60% reaction yield

Alkaline Battery Zn (s) + 2OH − (aq) → ZnO (s) + H 2 O (l) + 2e − [E° = 1.28 V] 2MnO 2(s) + H 2 O (l) + 2e − → Mn 2 O 3(s) + 2OH − (aq) [E° = V] Overall reaction: Zn (s) + 2MnO 2(s) ZnO (s) + Mn 2 O 3(s) [e° = 1.43 V]

Li Ion Battery The positive electrode half-reaction is: [45]half-reaction [45] The negative electrode half reaction is:

Redox Flow Battery New design

Vanadium Redox Flow Battery Aqueous system Discharge Positive Electrode: V (V) O H + + e - → V (IV) O 2+ + H 2 O (E 0 = 0.99 V vs. SHE) Negative Electrode: V 2+ → V 3+ + e - (E 0 = V vs. SHE) E cell =0.99 V – (-0.26) = 1.25 V

VRF Future Research Electrodes for fast e- transfer H 2 interference during charging cycle V 3+ + e - → V 2+ (E 0 = V vs. SHE) 2H + + 2e -  H 2 E0 = 0.00 V (preferred) Generates OH - poisons VRF battery