Question

7. Determine Eo for the following reaction, using the given standard reduction potentials: Cu(s) + Fe2+(aq) → Cu+(aq) + Fe(s)

Eo for Fe2+(aq) = -0.44 V

Eo for Cu+(aq) = 0.52 V

8. Consider the following half-reactions. Which of these is the strongest oxidizing agent listed here?

I2(s) + 2 e- → 2 I-(aq) Eo = 0.53 V

S2O82-(aq) + 2 e- → 2 SO42-(aq) Eo = 2.01 V

Cr2O72-(aq) + 14 H+ + 6 e- → Cr3+(aq) + 7 H2O(l) Eo = 1.33 V

1. I2(s)

2. I-(aq)

3. S2O82-(aq)

4. SO42-

5. Cr2O72-

6. Cr3+(aq)

Answer 1

Q7)

Reduction Reaction: Fe(2+)(aq) + 2e- --------------------- Fe(s), Eo(reduction) = -0.44V

Oxidation Reaction: Cu(s) ------------------ Cu+(aq) + e- , Eo(reduction) = -0.52V

Net Reaction:

Fe2+(aq) + 2Cu(s) ---------------------- Fe(s) + 2Cu+(aq)

Eo(cell) = Eo(oxidation) + Eo(reduction) = (-0.44) + (-0.52) = -0.96V

8)

The E0 reduction is highest for S2O8(2-) reaction with a value of E0 equal to 2.01V

Hence the strongest oxidizing agent will be S2O8(2-)

Correct answer is Option 3