Question

20#15

Given the following half-reactions and associated standard reduction potentials:
AuBr−4(aq)+3e−→Au(s)+4Br−(aq)
E∘red=−0.858V
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
IO−(aq)+H2O(l)+2e−→I−(aq)+2OH−(aq)
E∘red=+0.49V
Sn2+(aq)+2e−→Sn(s)
E∘red=−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.

E∘max = _____ 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.

E∘min = _____ V

Answer 1

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

AuBr−4(aq)+3e−→Au(s)+4Br−(aq) [Lowest]
E∘red=−0.858V
Eu3+(aq)+e−→Eu2+(aq)
E∘red=−0.43V
IO−(aq)+H2O(l)+2e−→I−(aq)+2OH−(aq) [Highest]
E∘red=+0.49V
Sn2+(aq)+2e−→Sn(s)
E∘red=−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)+2e−→I−(aq)+2OH−(aq) [Highest]
E∘red=+0.49V

Cathode : Sn2+(aq)+2e−→Sn(s)
E∘red=−0.14V

And anode : Eu3+(aq)+e−→Eu2+(aq)
E∘red=−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