help me solve G and H! i have wrote and done thee rest of the other parts. plz r

Question

help me solve G and H!

i have wrote and done thee rest of the other parts. plz refer to it if you need it :D

and plz avoid writing a single alphabet or something to joke around.. i would prefer my question to be not answers....:(

1) A student measures the potential of a cell made up with 1 M CuSO4 in one solution and 1 M AgNO3 in the other. There is a Cu electrode in the CuSo4 and an Ag electrode in the AgNO3, and the cell is set up in a "voltmeter." She finds that the potential, or voltage, of the cell, E^0cell, is 0.45 V, and that the Cu electrode is negative

a)At which electrode is oxidation occurring?

Cu

b)Write the equation for the oxidation reaction.

Cu ----> Cu2+ + 2e-

0.45

e) If E^0Ag+, Agred equals 0.80 V, as in standard tables of electrode potentials, what is the value of the potential of oxidation reaction of copper, E^0Cu, Cu2+oxid?

-0.35

f) Write the net ionic equation for the spontaneous reaction that occurs in teh cell that the student studied.

Cu + 2Ag^+ -------> Cu^2+ + 2Ag

g.) The student adds 6 M NH3 to be CuSO4 solution until the Cu2+ ion is essentially all converted to Cu(NH3)4 2+ ion. The voltage of the cell, Ecell, goes up to 0.92 V and the Cu electrode is still negative. Find the residual concentration of Cu 2+ ion in the cell. (Use eq.3)

Eq. Ecell = E0cell (- 0.0592/ 2) log ([Cu^2+]/[Ag^+]^2)                                                         ____?___M

h.) IN Part g, [Cu(NH3)4 2+] is about 0.05M, and [NH3] is about 3 M. Given those values and the result in Part 1g for [Cu 2+], calculate K for the reaction:

Cu(NH3)4^2+ (aq) <---> Cu^2+(aq) + 4(NH3) (aq)

                                                                                                                                   ______?______

Please let me know if I've done something wrong! Thank you!

Explanation / Answer

For given cell,

E0cell = +0.45 V, Ecell = 0.92 V

g) We have [Ag+] = [AgNO3] = 1 M, [Cu2+] = ?

Nernst equation gives the cell EMF as,

Ecell = E0cell – (0.0591/2) log ([Cu2+]/[Ag+]2)

Let us put these values in above Nernst equation,

0.92 = +0.45 – (0.0591/2) log ([Cu2+]/(1)2)

0.92 = +0.45 – (0.02955) log ([Cu2+])

(0.02955) log ([Cu2+]) = 0.45 – 0.92

(0.02955) log ([Cu2+]) = – 0.47

log ([Cu2+]) = –0.47/0.02955

log ([Cu2+]) = – 15.91

[Cu2+] = 10– 15.91

[Cu2+] = 1.24 x 10-16 M

========================================

h)

Cu(NH3)42+ (aq.) <-------------> Cu2+ (aq.) + 4 NH3 (aq.)

Equilibrium constant for this reaction will be given as,

K = {[Cu2+ (aq.)][NH3 (aq.)]2}/[Cu(NH3)42+ (aq.)]

We have given with, [Cu(NH3)42+] = 0.05 M, [NH3] = 3 M,

And we calculated, [Cu2+] = 1.24 x 10-16 M

Let us put all these values in equilibrium constant expression (K),

K = (1.24 x 10-16 M)(3)4/(0.05)

K = 2.01 x 10-13

Hence the equilibrium constant K.

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