Please provide a detailed solution and/or link. 1.) For a galvanic cell Cu|Cu2+

Question

Please provide a detailed solution and/or link.

1.) For a galvanic cell Cu|Cu2+ (0.10 M)||Fe2+ (0.10 M), Fe3+ (0.10 M)|Pt, (a) indicate the cathode and anode, (b) function of Pt, (c) cell’s potential, and (d) equilibrium constant of the cell reaction at 298 K given that E 0 (Cu2+/Cu) = 0.34 V, and E 0 (Fe3+ / Fe2+ ) = 0.77 V.

2.) For a potentiometric setup Ag/AgCl (3.0 M KCl)||ZnSO4 (x M)|Zn, the reference electrode Ag/AgCl (3.0 M KCl) has a constant potential of 0.20 V vs SHE, and E 0 [Zn2+/Zn] = -0.76 V vs SHE. What is the ZnSO4 concentration in the cell if the measured potential has a value of -1.2 V?

3.) For a redox reaction 2Mn2+ + 10CO2 + 8H2O = MnO4 - + 5C2O4 2- + 16H+ , calculate the Keq at 298 K given that:

MnO4 - + 8H+ +5e = Mn2+ + 4H2O E1^0 = 1.51 V vs SHE

C2O4 2- + 2e = 2CO2 E2^0 = 0.77 V vs SHE

Explanation / Answer

1.

(a0

Anode: Cu/Cu2+

Anode Half reacion :

Cu (s) ------------> Cu2+ (aq.) + 2 e-

Cathode : Pt, Fe2+/Fe3+ , Pt acts as inert electrode.

Cathode half reaction :

Fe3+ (aq.) + e- -----------> Fe2+ (aq.)

Balanced cell reaction :

Cu (s) + 2 Fe3+ (aq.) --------> Cu2+ (aq.) + 2 Fe2+ (aq.)

E0cell = E0Fe3+/Fe2+ - E0Cu2+/Cu = 0.77 - 0.34 = 0.43 V

Applying Nernst equation,

Ecell = E0cell - ( 0.05916 / n ) * ln{[Fe2+]2[Cu2+] / [Fe3+]2}

Ecell = 0.43 - ( 0.05916 / 2 ) * ln(0.102 * 0.10 / 0.102)

Now,

deltaG0 = - nFE0cell and deltaG0 = - R T lnK

SO,

R T lnK = n F E0cell

lnK = 2 * 96485 * 0.43 / ( 8.314 * 298 )

lnK = 33.5

K = 3.51 * 1014

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