L-Determine the Eo cell, the ?Go and the Equilibrium constant for the following

Question

L-Determine the Eo cell, the ?Go and the Equilibrium constant for the following reaction in standard conditions. Express if the battery is spontaneous. Reduction: Fe+3 +le Reduction Ni+2(aa) + 2e- ? Fea) E° red 0.77 volts Ni (s) Eo red = =-0.25 volts Spontaneity YesNo 2.- For the previous reaction determine the Cell potential if the [Fe ] 0.02 M, [Fe ] -0.01 M[ Ni2] 0.04 M and express is the battery is spontaneous. E° cell- 3.-a.- Determine the Mercury solubility at room temperature in Mercury(1) Sulfate Mercury solubility Spontaneity Yes No

Explanation / Answer

1. For balanced cell reaction,

Ni(s) + 2Fe3+(aq) --> Ni2+(aq) + 2Fe2+(aq)

Eocell = Ecathode - Eanode

          = 0.77 - (-0.25)

          = 1.02 V

Using,

dGo = -nFEocell

with,

n= 2

F = Faraday's constant

we get,

dGo = -2 x 96500 x 1.02/1000 = -196.860 kJ

Using,

dGo = -RTlnKc

with,

Kc = equilibrium constant

R = gas constant

T = 298 K

we get,

-196860 = -8.314 x 298 lnKc

Kc = 3.22 x 10^34

dGo = -ve value, thus spontaenity : Yes

2. For the concentrations,

[Fe3+] = 0.02 M

[Fe2+] = 0.01 M

[Ni2+] = 0.04 M

Kc = [Ni2+][Fe2+]^2/[Fe3+]^2

     = (0.04)(0.01)^2/(0.02)^2

     = 0.01

Feeding in dGo equation,

dGo = -RTlnKc

        = -8.314 x 298 ln(0.01)/1000

        = 11.41 kJ

Since sign of dGo = +ve

Spontaneity = No

3.

Ksp Hg2SO4 = 6.80 x 10^-7

Ksp = [Hg+]^2.[SO4^2-]

let x amount of salt is in solution at room temperature. Then 2x amount of Hg+ and 2 amount of SO4^2- would be in solution

Ksp = (2x)^2.(x)

6.80 x 10^-7 = (2x)^2.(x)

solubility fo mercury = x = 5.54 x 10^-3 M

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