b)Same as part a but with copper and lead. c) Same as part a but with cadmium an

Question

b)Same as part a but with copper and lead.

c) Same as part a but with cadmium and iron.

Standard Electrode Potential, V(V) Electrode Reaction +1.420 +1.229 Pt2+ + 2e-Pt Ag+ + e-Ag +0.800 +0.771 +0.401 +0.340 0.000 -0.126 -0.136 -0.250 -0.277 -0.403 -0.440 -0.744 -0.763 -1.662 -2.363 -2.714 -2.924 Increasingly inert (cathodie) Og + 2H2O + 4e"- 4(OH-) Cu2+ + 2e-Cu 2H+ + 2e- H2 Pb2+ + 2e-Pb Ni2+ + 2e- Ni Co2+ + 2eCo Cd2+ + 2e-Cd Fe2+ + 2e-Fe Increasingly active (anodic) Zn2+ + 2e- Zn AP+ + 3e-Al Mg2+ + 2e- Mg Na+ + e- Na The voltages in Table 17.1 are for the half-reactions as reduction reactions, with the electrons on the left-hand side of the chemical equation; for oxidation, the direction of the reaction is reversed and the sign of the voltage changed. Consider the generalized reactions involving the oxidation of metal M, and the reduction of metal M2 as Vs (17.16a) (17.16b) where the V's are the standard potentials as taken from the standard emf series. Because metal Mi is oxidized, the sign of V is opposite to that as it appears in Table 17.1 Addition of Equations 17.16a and 17.16b yields (17.17) nd the overall cell potential is

Explanation / Answer

we know that

the half cell with higher reduction potential acts as cathode

and

the one with lower reduction potential acts as cathode

now

Eo cell = Eo cathode -Eo anode

1) copper and zinc

Eo for copper is 0.34

Eo for zinc is -0.763

so

copper is cathode and zinc is anode

so

Eo cell = 0.34 + 0.763 = 1.103 V

so

the net potential is 1.103 V

2)

copper and zinc

Eo for copper is 0.34

Eo for lead is -0.126

so

copper is cathode and lead is cathode

so

Eo cell = 0.34 + 0.126 = 0.466 V

so

the net potential is 0.466 V

3) cadmium and iron

Eo for cadmium is -0.403

Eo for iron is -0.44

so

cadmium is catode and iron is anode

so

Eo cell = -0.403 + 0.44 = 0.037 V

so

the net potential is 0.037 V

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