Please solve all of question 1(parts a through g). I really need help as soon as

Question

Please solve all of question 1(parts a through g). I really need help as soon as possible! I. Consult the table of reduction potentials (pages 123-124 in the lab manual) to help you answer this question. Suppose that each combination of half-cells listed below is con- nected with a salt bridge and allowed to react spontaneously. For each cell, (i) write the reduction half-reaction and the oxidation half-reaction, then (i) combine those half-re- actions into the overall balanced redox reaction for the cell. Finally, (iii) use the reduction potentials from the table to calculate the cell potential under standard conditions. Show your work clearly! a. Cu/Cu and Ni/Ni half-cells b. Ag/Ag' and Au/Au half-cells c. Pb/Pb and Al/AP+ half-cells Continued on next page.

Explanation / Answer

Q1

Remember that each species will have a specific reduction potential. Remember that this is, as the name implies, a potential to reduce. We use it to compare it (numerical) with other species.

Note that the basis if 2H+ + 2e- -> H2(g) reduction. Therefore E° = 0 V

All other samples are based on this reference.

Find the Reduction Potential of each reaction (Tables)

Cathode

Anode

The most positive has more potential to reduce, it will be reduced

The most negative will be oxidized, since it will donate it selectrons

For total E°cell potential:

E°cell = Ered – Eox

Eox = -Ered of the one being oxidized

a

Ni2+ + 2 e Ni(s) 0.25

Cu2+ + 2 e Cu(s) +0.337

Clearly, Cu will reduce, cathode, and Ni will oxidize, anode

E°cell = Ecathode - Eanode = 0.337 -- 0.25 = 0.587 V

b)

Ag+ + e Ag(s) +0.7996

Au3+ + 3 e Au(s) +1.52

Au reduces, cathode

Ag oxidizes, anode

E°cell = Ered - Eox =1.52 - 0.7996

E°cell = 0.7204 V

c)

Pb2+ + 2 e Pb(s) 0.126

Al3+ + 3 e Al(s) 1.662

Aluminium oxidizes, Lead reduces

so

E°cell = -0.126 - -1.662

E°cell = 1.536 V

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