PLEASE Really need help in figuring this out Ag+ (0.010M) + e- -> Ag(s) E^0 = +0

Question

PLEASE Really need help in figuring this out

Ag+ (0.010M) + e- -> Ag(s) E^0 = +0.80V

Cr3+ (0.010M) + 3e- -> Cr(s) E^0= -0.74V

1. Write the equation for the half-reaction occuring at the cathode

2. Write equation for half-rxn occuring at anode

3. Write te equation for the cell reaction

4. What is the standard cell potential, E^0 cell, for the cell?

5. Realizing the nonstandard concentrations, what is the actual potential, Ecell, for the cell? Hint: What is the value of n in the Nernst equation.

Explanation / Answer

1.) Ag+ + e- --> Ag
The cathode is where reduction occurs. The lower the value of the standard reduction potential of a half-cell reaction, the more likely it is to be reduced. Because -0.74 V (Cr) is less than 0.80 V (Ag), Crwill be oxidized (As is seen in part 2)--And therefore, silver will be reduced.

2) Cr--> Cr3+ + 3e-

3.) 3Ag+ + Cr --> Cr3+ + 3Ag (simply balance charge)

4.) At standard conditions, Ecell = Ecathode - Eanode
Ecell = 0.80 V - (-0.74) V = 1.54 V

5)

E cathode = 0.80 V - 0.0592/3* log(1/[0.010]^3) = 0.682 V
E anode = -0.74 V - 0.0592/1*log(1/[0.0010]) = -0.8584 V
Ecell = 0.682 V - (-0.8584 V) = 1.5404V

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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