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

ID: 1044726 • Letter: L

Question

L-Determine the Eo cell, the ?G° and the Equilibrium constant for the following reaction in standard conditions. Express if the battery is spontaneous. Reduction: Fe+3 ? + 1e- ? Fe+2(aa) ered 0.77 volts Reduction Ni+2(aq) +2e-? Ni (s) Eored..-Q25 volts K ?Spontaneity Yes, No 2.- For the previous reaction determine the Cell potential if the [Fe"] 0.02 M, [Fe]-0.01 MINi]- 0.04 M and express is the battery is spontaneous. Spontaneity Yes_ . No = 3.-a.- Determine the Mercury solubility at room temperature in Mercury() Sulfate Mercury solubility 4.- Determine if the solubility of Mercury will diminish or increase when Mercury (I) Sulfate is dissolved in a solution of 0.001 M Sodium sulfate. The solubility will diminish 5.- Explain why a colloidal dispersion will sediment at different rates if instead of adding a Ba Cl increase Aluminum hydroxide is added. Which one will sediment faster? The colloidal dispersion will sediment faster with Ba Cl2 or Al(OH)

Explanation / Answer

1. For the given cell

Eocell = Ecathode - Eanode

           = 0.77 - (-0.25)

           = 1.02 V

dGo = -nFEo

with,

F = Faraday's constant

n = 2

we get,

dGo = -2 x 96500 x 1.02/1000 = -196.86 kJ/mol

and,

dGo = -RTlnK

with,

K = equilibrium constant

R = gas constant

T = 298 K

we get,

-196860 = -8.314 x 298 lnK

K = 3.22 x 10^34

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2. when,

[Fe3+] = 0.02 M

[Fe2+] = 0.01 M

[Ni2+] = 0.04 M

Eocell = 1.02 V

dGo = -RTlnK

        = -8.314 x 298 ln((0.01)^2.(0.04)/(0.02)^2)/1000

        = 11.41 kJ/mol

Since dGo is +ve, the reaction is non-spontaneous

Spontaneity : No

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3. Ksp of Hg2SO4 = [Hg+]^2.[SO4^2-] = 6.80 x 10^-7

3.

without sulfate added

with x amount of Hg2SO4 in solution

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

solubility of mercury = cube root(6.80 x 10^-7/4) = 5.54 x 10^-3 M

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4. After 0.001 M Na2SO4 added

[SO4^2-] = 0.001 M

So,

solubility of mercury = [Hg+] = sq.rt.(Ksp/[SO4^2-])

                                              = sq.rt.(6.80 x 10^-7/0.001)

                                              = 0.0261 M

The solubility will increase after addition of Na2SO4

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