Question

Using the information in the table:

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.

Which combination of metals, if used to create an electrochemical cell, would produce the largest voltage?

Answer 1

We need to create an electro-chemical cell of Zinc and Silver

Oxidation Half Cell :

Zn(s) → Zn2+ (aq) + 2e

Reduction Half Cell :

Ag+ (aq) + e + Ag(s)

Overall reaction will be

2 Ag+ (aq) + Zn(s) + 2Ag(s) + Zn2+ (aq)

E'(run) = E'oxidation) + E'(reduction) = 0.76 +0.80 = 1.56V

Note - Post any doubts/queries in comments section. Rate the answer positive if found helpful.