20#15 Given the following half-reactions and associated standard reduction poten

Question

20#15

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)+2eI(aq)+2OH(aq)
Ered=+0.49V
Sn2+(aq)+2eSn(s)
Ered=0.14V

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

b) Calculate the value of this emf.

Emax = _____ V

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

d) Calculate the value of this emf.

Emin = _____ V

Explanation / Answer

a) the highest positive emf will be for the combination where

anode has lowest standard reduction potential

Cathode has highest standard reduction potential

Let us check the reduction potential values

AuBr4(aq)+3eAu(s)+4Br(aq) [Lowest]
Ered=0.858V
Eu3+(aq)+eEu2+(aq)
Ered=0.43V
IO(aq)+H2O(l)+2eI(aq)+2OH(aq) [Highest]
Ered=+0.49V
Sn2+(aq)+2eSn(s)
Ered=0.14V

So the cell reaction will be

2Au(s) + 8Br^ - +3IO^ -(aq) + 3H2O(l) --> 2AuBr4^ - (aq) + 3I^ - (aq) + 6OH^ - (aq)

b) the emf will be

E0cell = E0cathode - E0anode = 0.49 - (-0.858) = 1.348 V

c) For lowest positive emf value, the reduction potential difference should be minimum

We can not take

IO(aq)+H2O(l)+2eI(aq)+2OH(aq) [Highest]
Ered=+0.49V

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

And anode : Eu3+(aq)+eEu2+(aq)
Ered=0.43V

The reaction will be

2Eu^2+(aq)+Sn^2+(aq)2Eu^3+(aq)+Sn(s)

d) emf of cell

E0cell = E0cathode - E0anode= -0.14 - (-0.43) = +0.29 V

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