Question

Using the table below:

Liu lur the reaction between Zn and Cu2+ ions is 1.1030 V, we can use the known value for the half-cell potential for zinc to determine the half-cell potential for copper: Zn(s) → Zn2+(aq) + 2e + Cu2+(aq) + 2e → Cu(s) Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s) E half-cell = 0.7628 V Eºhalf-cell = 0.3402 V Eºcell = 1.1030 V In this manner we can determine the relative cell potentials for any series of half-cells, such as those given in the following table: Half-reaction E° (V) Ag+ (aq) + e + Ag(s) 0.80 Fe3+(aq) + 1e → Fe2+(aq) 0.77 Cu2+(aq) + 2e → Cu(s) 0.34 2H+(aq) + 2e → H2(g) 0.00 Pb2+(aq) + 2e → Pb(s) -0.13 Ni2+(aq) + 2e → Ni(s) -0.25 Cd2+(aq) + 2e → Cd(s) -0.40 -0.44 Fe2+(aq) + 2e → Fe(s) -0.76 Zn2+(aq) + 2e → Zn(s) Sle 5-1: Half-cell potentials. All potentials are tabulated as reduction potentials. Standard uction potentials refer to measurements taken in the standard state, i.e. 1 M concentrations all ions, 25°C, and 1 atm of pressure for all gases.

19. Three combinations of metals are listed below, which combination would produce the largest voltage if they were used to construct an electrochemical cell?

		Copper (Cu) with zinc (Zn)
		Lead (Pb) with zinc (Zn)
		Lead (Pb) with cadmium (Cd)

Answer 1

Cell: copper (c) and Zinc (zn) Reduction potential given! - E cut?lcu = 0.34 V. Ernt2/zn = -0.760 The metal which has more reduction potential, ie it was more tendency to undency to getting getting reales reduced in that metal will act as cathode and the other metal will behave as anode. i cu cathode In Anode. Anode :- cathode :- n n + 2e Ern/20+2 = 0.76V cut2 +2e- u E cut2/4 = 0.34 2015) + cut? (aq) - znt? (9) + Cu(s). Ecell- Eżnienta & Ecure = 0.76 +0.34 = 110v cell. 2.! Lead (Pb) and zinc (zn.) Reduction potential - Epota/Pb = -0.13v Ezut2/ zn =-0.76 V 2. Pb Cathode (Because Pb has more Reduction porenti zn Anode. Anode : znits) ant² (aq) +2e - Eznlar² = 0.760 Cathode :- Poth cap) + ae- Pbls) Ept2lPb = -0.13V zöls) +p6409) zot kaq) + Phts) Filll - Een/zit2 + Epht? / Pb = 0.76-0.13 = 0.63 V