Given the following half-reactions and associated standard reduction potentials:

Question

Given the following half-reactions and associated standard reduction potentials: AuBr4(aq)+3eAu(s)+4Br(aq) Ered=0.858V
Eu3+(aq)+eEu2+(aq)
Ered=0.43V
IO(aq)+H2O(l)+2e I(aq)+2OH(aq)
Ered=+0.49V
Sn2+(aq)+2eSn(s)
Ered=0.14V

Part A
Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf.

Part B

Calculate the value of this emf.

Part C

Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf.

Express your answer as a chemical equation. Identify all of the phases in your answer.

Part D

Calculate the value of this emf.

Express your answer using two decimal places.

Please help!! So stuck!

Explanation / Answer

A) Write the cell reaction for the combination of these half-cell reactions that leads to the largest positive cell emf.

For lead having the greatest value, we need the lowest value + lead value

Pb2+ + 2 e Pb(s) 0.126

then, choose a LOWER potential

Best option will be:

AuBr4(aq)+3eAu(s)+4Br(aq) Ered=0.858V

overal reaction:

3Pb2+ + 6e 3Pb(s) 0.126

2AuBr4(aq)+6e2Au(s)+8Br(aq) Ered=0.858V

inver oxidation

3Pb2+ + 6e 3Pb(s) 0.126

2Au(s)+8Br(aq)2AuBr4(aq)+6e Ered=0.858V

B)

E º = E reduction - E oxidation

E º = -0.126 - - 0.858 = 0.732 V

C)

Smallest positive potential

the potential mst be < than Lead, in order to satisfy positive value,

must be nearest to the E red value of Pb

so

Pb2+ + 2 e Pb(s) 0.126

Sn2+(aq)+2eSn(s) Ered=0.14V

D)

E º = E red - Eox = -0.126 - -0.14 = 0.014 V

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