1. www.soran.edu.iq Inorganic chemistry Assistance Lecturer Amjad Ahmed Jumaa  Batteries and their application.  Primary (nonrecharge able) batteries.  Flow batteries (fuel cells). 1

2. www.soran.edu.iq Solution: Dividing the reaction into half-reactions; Br 2 (aq) + 2e - → 2Br - (aq) Eº unknown. =Eº bromine =? V Zn(s) → Zn +2 (aq) + 2e - Eº= -0.76V. Calculating Eº bromine. : Eº cell. = Eº cathode. - Eº anode. Eº cell. = Eº bromine. - Eº zinc. Thus, Eº bromine. = Eº cell. + Eº zinc. = 1.83V + (- 0.76V) = 1.07V. Batteries and their application: Primary (non recharge able) batteries: (a) Dry cell: A zinc anode in the form of a can houses a mixture of MnO 2, and an electrolyte paste, consisting of NH 4 Cl, ZnCl 2, H 2 O, and starch. Powdered graphic improves conductivity; the cathode is in active graphite rod.

3. www.soran.edu.iq Anode (oxidation): Zn(s) → Zn +2 (aq) + 2e -. Cathode (reduction): The cathode half-reaction is complex and not completely understood. 2MnO 2 (s) +2NH 4 + (aq) +2e - → Mn 2 O 3 (s) +2NH 3 (aq) +H 2 O (l) Zn +2 (aq) +2NH 3 (aq) +2Cl - (aq) → Zn (NH 3 ) 2 Cl 2 (s). Overall cell reaction: 2MnO 2 (s) + 2NH 4 Cl (aq) +Zn(s) → Zn (NH 3 ) 2 Cl 2 (s) +H 2 O (l) +Mn 2 O 3 (s). E cell = 1.5V.

4. www.soran.edu.iq (b) Alkaline battery: The alkaline battery is an improved dry cell. The half-reactions are essentially the same, but electrolyte is a KOH paste. Anode (oxidation): Zn(s) + 2OH - (aq) → ZnO(s) +H 2 O (l) +2e -. Cathode (reduction): MnO 2 (s) +2H 2 O (l) +2e - → Mn (OH) 2 (s) + 2OH - (aq). Overall cell reaction: Zn(s) + MnO 2 (s) + H 2 O (l) → ZnO(s) + Mn (OH) 2 (s). E cell. =1.5V. Anode (oxidation): Zn(s) + 2OH - (aq) → ZnO(s) +H 2 O (l) +2e -. Cathode (reduction): MnO 2 (s) +2H 2 O (l) +2e - → Mn (OH) 2 (s) + 2OH - (aq). Overall cell reaction: Zn(s) + MnO 2 (s) + H 2 O (l) → ZnO(s) + Mn (OH) 2 (s). E cell. =1.5V.

5. www.soran.edu.iq Secondary (recharge able) batteries: Lead-acid battery: A typical 12-V lead-acid car battery has six cells connected in series. The anode is spongy, powdered Pb, and the cathode is powdered PbO 2. The grids are immersed in an electrolyte solution ~ 4.5 M H 2 SO 4. Anode (oxidation): Pb(s) + HSO 4 - (aq) → PbSO 4 (s) +H + +2e -.

6. www.soran.edu.iq Cathode (reduction): PbO 2 (s) + 3H + (aq) + HSO 4 - (aq) +3e - → PbSO 4 (s) +2H 2 O (l). Both half-reaction produce Pb +2, one through oxidation of Pb, the other through reduction of PbO 2, Overall cell-reaction (discharge): PbO 2 (s) +Pb(s) + 2H 2 SO 4 (aq) → 2PbSO 4 (s) + 2H 2 O (l). E cell. =2.1V. When the (cell recharges), it uses electrical energy as an electrolytic cell, and the half-cell and overall reactions are reversed: Overall cell-reaction (recharge): 2PbSO 4 (s) + 2H 2 O (l) → PbO 2 (s) + Pb(s) +2H 2 SO 4 (aq). 17

7. www.soran.edu.iq Flow batteries (fuel cells): A fuel cell being designed for car oxidizes H 2 to H 2 O. At the catalyst layer (platinum) in the anode, H 2 is split, and the electron enters the circuit. At the catalyst layer in the cathode, O 2 is split, reduced, and combined with H+. Anode (oxidation): H 2 (g) → 2H + (aq) +2e - Cathode (reduction): 1/2 O 2 (g) +2H + (aq) +2e - → H 2 O (g) Overall cell-reaction: H 2 (g) +1/2 O 2 (g) → H 2 O (g) E cell. =1.2V. 18