given the following half reactions and associated standard reduction potentials:

Question

given the following half reactions and associated standard reduction potentials:

AuBr4^-(aq)+3e^- ----->Au(s)+4Br^-(aq)

Ered=-0.858V

Eu^3+(aq)+e^- ------>Eu^2+(aq)

Ered=-0.43V

IO^-(aq)+H2O(l)+2e^-=I^-+2OH^-(aq)

Ered=+0.49V

Sn^2+(aq)+2e- ----->Sn(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. Express your answer as a chemical equation. Identify all phases

b)calculate the value of this emf

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

d)calculate the value of this emf

Explanation / Answer

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

(b) Eo(cell) = Ered(IO3-/I-) - Eo(AuBr4-/Au)

= 0.49 - (-0.858) = 1.348 V

(c) Sn^2+(aq) + 2 Eu^2+(aq) => Sn(s) + Eu^3+(aq)

(d) Eo(cell) = Ered(Sn2+/Sn) - Eo(Eu3+/Eu)

= -0.14 - (-0.43) = 0.29 V

Related Questions

Navigate

• Browse (All)
• Browse G
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.