Chemsheets AS006 (Electron arrangement) 14/04/2017 www.CHEMSHEETS.co.uk ELECTROCHEMISTRY © www.chemsheets.co.uk A2 046 20-Jul-12

Zn2+(aq) + 2 e–  Zn(s)

Zn  Zn2+ + 2 e- oxidation Cu2+ + 2 e-  Cu reduction - electrode anode oxidation + electrode cathode reduction electron flow At this electrode the metal loses electrons and so is oxidised to metal ions. These electrons make the electrode negative. At this electrode the metal ions gain electrons and so is reduced to metal atoms. As electrons are used up, this makes the electrode positive. Zn Cu Zn  Zn2+ + 2 e- oxidation Cu2+ + 2 e-  Cu reduction

Standard Conditions Concentration 1.0 mol dm-3 (ions involved in ½ equation) Temperature 298 K Pressure 100 kPa (if gases involved in ½ equation) Current Zero (use high resistance voltmeter)

S tandard H ydrogen E lectrode

Emf = E = Eright - Eleft

Pt(s) | H2(g) | H+(aq) || Cu2+(aq) | Cu(s)

ROOR Ni(s) | Ni2+(aq) || Sn4+(aq), Sn2+(aq) | Pt(s) K(s) | K+(aq) || Mg2+(aq) | Mg(s)

ELECTRODE POTENTIALS – Q1 Emf = Eright - Eleft - 2.71 = Eright - 0 Eright = - 2.71 V

ELECTRODE POTENTIALS – Q2 Emf = Eright - Eleft Emf = - 0.44 - 0.22 Emf = - 0.66 V

ELECTRODE POTENTIALS – Q3 Emf = Eright - Eleft Emf = - 0.13 - (-0.76) Emf = + 0.63 V

ELECTRODE POTENTIALS – Q4 Emf = Eright - Eleft +1.02 = +1.36 - Eleft Eleft = + 1.36 - 1.02 = +0.34 V

ELECTRODE POTENTIALS – Q5 Emf = Eright - Eleft a) Emf = + 0.15 - (-0.25) = +0.40 V b) Emf = + 0.80 - 0.54 = +0.26 V c) Emf = + 1.07 - 1.36 = - 0.29 V

ELECTRODE POTENTIALS – Q6 Emf = Eright - Eleft a) Eright = +2.00 - 2.38 = - 0.38 V Ti3+(aq) + e-  Ti2+(aq) b) Eleft = -2.38 - 0.54 = - 2.92 V K+(aq) + e-  K(aq) c) Eright = - 3.19 + 0.27 = - 2.92 V Ti3+(aq) + e-  Ti2+(aq)

ELECTRODE POTENTIALS – Q7 a) Cr(s) | Cr2+(aq) || Zn2+(aq) | Zn(s) Emf = -0.76 - (-0.91) = +0.15 V b) Cu(s) |Cu2+(aq)|| Fe3+(aq),Fe2+(aq)| Pt(s) Emf = +0.77 - 0.34 = +0.43 V c) Pt(s) | Cl-(aq)| Cl2(g) || MnO4-(aq),H+(aq),Mn2+(aq)| Pt(s) Emf = +1.51 – 1.36 = +0.15 V

The more +ve electrode gains electrons (+ charge attracts electrons)

– + e– + 1.10 V + 0.34 V – 0.76 V Cu2+ + Zn → Cu + Zn2+ + –ve electrode +ve electrode e– + 1.10 V + 0.34 V Cu2+ + 2 e-  Cu – 0.76 V Zn2+ + 2 e-  Zn Cu2+ + Zn → Cu + Zn2+

PREDICTING REDOX REACTIONS – Q1 + –ve electrode +ve electrode e– + 0.51 V – 0.25 V Ni2+ + 2 e-  Ni – 0.76 V Zn2+ + 2 e-  Zn Ni2+ + Zn → Ni + Zn2+

PREDICTING REDOX REACTIONS – Q2 + –ve electrode +ve electrode e– + 0.46 V + 0.80 V Ag+ + e-  Ag + 0.34 V Cu2+ + 2 e-  Cu 2 Ag+ + Cu → 2 Ag + Cu2+

PREDICTING REDOX REACTIONS – Q3 a Mg(s)|Mg2+(aq)||V3+(aq),V2+(aq)|Pt(s) – –ve electrode +ve electrode e– + 2.10 V – 0.26 V V3+ + e-  V2+ – 2.36 V Mg2+ + 2 e-  Mg YES: Mg reduces V3+ to V2+

PREDICTING REDOX REACTIONS – Q3 b + –ve electrode +ve electrode e– + 0.59 V + 1.36 V Cl2 + 2 e-  2 Cl- + 0.77 V Fe3+ + e-  Fe2+ NO: Cl- won’t reduce Fe3+ to Fe2+

PREDICTING REDOX REACTIONS – Q3 c Pt(s)|Br-(aq),Br2(aq)||Cl2(g)|Cl-(aq)|Pt(s) + –ve electrode +ve electrode e– + 0.27 V + 1.36 V Cl2 + 2 e-  2 Cl- + 1.09 V Br2 + 2 e-  2 Br- YES: Cl2 oxidises Br- to Br2

PREDICTING REDOX REACTIONS – Q3 d Sn(s)|Sn2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) – + –ve electrode +ve electrode e– + 0.91 V + 0.77 V Fe3+ + e-  Fe2+ – 0.14 V Sn2+ + 2 e-  Sn YES: Sn reduces Fe3+ to Fe2+

PREDICTING REDOX REACTIONS – Q3 e + –ve electrode +ve electrode e– + 0.03 V + 1.36 V + 1.33 V Cl2 + 2 e-  2 Cl- Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O NO: H+/Cr2O72- won’t oxidise Cl- to Cl2

PREDICTING REDOX REACTIONS – Q3 f Pt(s)|Cl-(aq)|Cl2(g)||MnO4- (aq),H+(aq),Mn2+(aq)|Pt(s) + –ve electrode +ve electrode e– + 0.03 V + 1.51 V + 1.36 V MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2O Cl2 + 2 e-  2 Cl- YES: H+/MnO4- oxidises Cl- to Cl2

PREDICTING REDOX REACTIONS – Q3 g Fe(s)|Fe2+(aq)||H+(aq)|H2(g)|Pt(s) – –ve electrode +ve electrode e– + 0.44 V 0.00 V 2 H+ + 2 e-  H2 – 0.44 V Fe2+ + 2 e-  Fe YES: H+ oxidises Fe to Fe2+

PREDICTING REDOX REACTIONS – Q3 h + –ve electrode +ve electrode e– + 0.34 V + 0.34 V Cu2+ + 2 e-  Cu 0.00 V 2 H+ + 2 e-  H2 NO: H+ won’t oxidise Cu to Cu2+

PREDICTING REDOX REACTIONS – Q4 + + 1.36 V Cl2 + 2 e-  2 Cl- + 0.77 V + 1.51 V Fe3+ + e-  Fe2+ MnO4- + 8 H+ + 5 e-  Mn2+ + 4 H2O NO + 1.33 V YES Cr2O72- + 14 H+ + 6 e-  2 Cr3+ + 7 H2O NO

PREDICTING REDOX REACTIONS – Q5a + –ve electrode 2.19 = 0.34 - Eleft +ve electrode Eleft = 0.34 – 2.19 = – 1.85 V e– + 2.19 V + 0.34 V Cu2+ + 2 e-  Cu ? V Be2+ + 2 e-  Be Be + Cu2+ → Be2+ + Cu

PREDICTING REDOX REACTIONS – Q5b –ve electrode When using SHE +ve electrode E = cell emf = – 1.90 V e– 1.90 V + 0.00 V 2 H+ + 2 e-  H2 ? V Th4+ + 4 e-  Th 4 H+ + Th → 2 H2 + Th4+

PREDICTING REDOX REACTIONS – Q6a Pt(s)|H2(g)|H+(aq)||Br2(aq),Br-(aq)|Pt(s) + –ve electrode +ve electrode e– + 1.09 V + 1.09 V Br2 + 2 e-  2 Br- 0.00 V 2 H+ + 2 e-  H2 H2 + Br2 → 2 H+ + 2 Br-

PPT - Electrode potentials PREDICTING REDOX REACTIONS – Q6b 14/04/2017 Cu(s)|Cu2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) + –ve electrode +ve electrode e– + 0.43 V + 0.77 V Fe3+ + e-  Fe2+ + 0.34 V Cu2+ + 2 e-  Cu 2 Fe3+ + Cu → 2 Fe2+ + Cu2+

Electrochemical cells i appreciate that electrochemical cells can be used as a commercial source of electrical energy j appreciate that cells can be non-rechargeable (irreversible), rechargeable and fuel cells k be able to use given electrode data to deduce the reactions occurring in non-rechargeable and rechargeable cells and to deduce the e.m.f. of a cell l understand the electrode reactions of a hydrogen-oxygen fuel cell and appreciate that a fuel cell does not need to be electrically recharged m appreciate the benefits and risks to society associated with the use of these cells

Non-rechargeable (primary) cells – Zinc-carbon -0.80 V Zn(NH3)22+ + 2 e-  Zn + 2 NH3 +0.70 V 2 MnO2 + 2 H+ + 2 e-  Mn2O3 + H2O Standard cell Short life Determine: a) cell emf b) overall reaction during discharge

Non-rechargeable (primary) cells – alkaline -0.76 V Zn2+ + 2 e-  Zn +0.84 V MnO2 + H2O + e-  MnO(OH) + OH- Determine: a) cell emf b) overall reaction during discharge Longer life

Non-rechargeable (primary) cells – lithium Very long life High voltage Determine: a) cell emf b) overall reaction during discharge

Rechargeable (secondary) cells In non-rechargeable (primary) cells, the chemicals are used up so the voltage drops In rechargeable (secondary) cells the reactions are reversible – they are reversed by applying an external current. It is important that the products from the forward reaction stick to the electrodes and are not dispersed into the electrolyte.

Rechargeable (secondary) cells – Li ion +0.60 V Li+ + CoO2 + e-  LiCoO2 -3.00 V Li+ + e-  Li Rechargeable Most common rechargeable cell Determine: a) cell emf b) overall reaction during discharge c) overall reaction during re-charge

Rechargeable (secondary) cells – lead-acid +1.68 V PbO2 + 3 H+ + HSO4- + 2 e-  PbSO4 + 2 H2O -0.36 V PbSO4 + H+ + 2 e-  Pb + HSO4- Determine: a) cell emf b) overall reaction during discharge c) overall reaction during re-charge Used in sealed car batteries (6 cells giving about 12 V overall)

Rechargeable (secondary) cells – nickel-cadmium +0.52 V NiO(OH) + 2 H2O + 2 e-  Ni(OH)2 + 2 OH- -0.88 V Cd(OH)2 + 2 e-  Cd + 2 OH- Determine: a) cell emf b) overall reaction during discharge c) overall reaction during re-charge

FUEL CELLS High efficiency (more efficient than burning hydrogen) How is H2 made? Input of H2/O2 to replenish so no need to recharge +0.40 V O2 + 2 H2O + 4 e-  4 OH- -0.83 V 2 H2O + 2 e-  H2 + 2 OH- Determine: a) cell emf b) overall reaction

From wikipedia (public domain)

Pros & cons of cells + portable source of electricity Pros & cons of non-rechargeable cells + cheap, small – waste issues Pros & cons of rechargeable cells + less waste, cheaper in long run – still some waste issues Pros & cons of fuel cells + water is only product – most H2 is made using fossil fuels, fuels cells expensive