± Cell Potential and Equilibrium 18 of 28 Constants | Periodic Table The equilib

Question

± Cell Potential and Equilibrium
18 of 28
Constants | Periodic Table
The equilibrium constant, K, for a redox reaction is related to the standard potential, E?, by the equation
lnK=nFE?RT
where n is the number of moles of electrons transferred, F (the Faraday constant) is equal to 96,500 C/(mol e?) , R (the gas constant) is equal to 8.314 J/(mol?K) , and T is the Kelvin temperature.
Standard reduction potentials
Reduction half-reaction   E? (V)
Ag+(aq)+e??Ag(s)   0.80
Cu2+(aq)+2e??Cu(s)   0.34
Sn4+(aq)+4e??Sn(s)   0.15
2H+(aq)+2e??H2(g)   0
Ni2+(aq)+2e??Ni(s)   ?0.26
Fe2+(aq)+2e??Fe(s)   ?0.45
Zn2+(aq)+2e??Zn(s)   ?0.76
Al3+(aq)+3e??Al(s)   ?1.66
Mg2+(aq)+2e??Mg(s)   ?2.37

Part A Part complete
Use the table of standard reduction potentials given above to calculate the equilibrium constant at standard temperature (25 ?C) for the following reaction:
Fe(s)+Ni2+(aq)?Fe2+(aq)+Ni(s)
Express your answer numerically.
View Available Hint(s)
K =    2.68×10^6 Correct
When E?>0 and K>1 the reaction favors the products.

Part B
Calculate the standard cell potential (E?) for the reaction
X(s)+Y+(aq)?X+(aq)+Y(s)
if K = 7.87×10?3.

Express your answer to three significant figures and include the appropriate units.

E? = ?

Explanation / Answer

part B)

X (s) + Y+ (aq) ---------------------> X+ (aq) + Y (s)

number of electrons transfered = 1

K = 7.87 x 10^-3

delta G   = - RT ln K

                 = -8.314 x 10^-3 x 298 x ln (7.87 x 10^-3)

                = 12.00 kJ / mol

delta G = - n F Eo

12.00 = - 1 x 96.5 Eo

Eo = - 0.124 V

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

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