Please help Using the following information calculate E degree for the reduction
ID: 1001717 • Letter: P
Question
Please help
Using the following information calculate E degree for the reduction reaction in Equation 5 (F = 96487 C mol^-1). Equation 5: MnO_3(s) + 2H^+ + 2e^- = 2MnO(s) + H_2O Species: MnO H_2O Mn_2O_3 H^+ delta G degree kJ/mol: -403 -237 -991 0 Starting from the general Equations 6 and 7, calculate the value of E at pH = 2 for the reduction described in Equation 5. (2.303RT/F = 0.059 V at 298 K). Equation 6: yOx + ne^- = x Red Equation 7: E = E degree - (RT/nF) Ln{(a_Red)^x/(a_Ox)^y} Using the solubility diagram of the sulfide of M(II) shown below, predict the concentration of M(II) in solution at pH 6. If the pH was changed to 7 what percentage of M(II) will be removed from the solution as the insoluble sulfide of M(II).Explanation / Answer
a)
Given balanced redox reaction is,
Mn2O3 (s) + 2H+ (aq.) + 2e– -----------> 2MnO (s) + H2O
We are supplied with the standard Gibbs energies of formation of species involved in above redox reaction. Let us find out Standard Gibbs energy change (GoRXN) of the reaction from them.
We know that,
Gibbs energy change of reaction(GoRXN) = Gfo(Product) – Gfo(Reactants)
i.e. (GoRXN) = [2 x Gfo(MnO) + Gfo(H2O)] – [Gfo(Mn2O3) + 2 Gfo(H+)]
(GoRXN) = [(2 x –403) + (–237)] – [(–991) + 2 (0)]
(GoRXN) = – 52 kJ
(GoRXN) = – 52000 J
Standard Gibbs energy change for reaction (GoRXN) and Standard electrode potential E0 is related as,
(GoRXN) = – nFE0 ……………..(1)
Where,
n = Number of electrons transferred = 2
F = Faraday’s constant = 96487 C/mol
(GoRXN) = – 52000 J calculated
E0 = ? want to find.
Let us put all the known values in eq. (1) and solve it for E0.
– 52000 = – 2 x 96487 x E0
52000 = 2 x 96487 x E0
52000 = 192974 x E0
E0 = 52000/192974
E0 = + 0.269 V
Standard electrode potential is 0.269 V.
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b)
Using concept in equation 6 we can have Expression for the EMF of the cell E for given redox reaction,
E = E0 – (RT/nF) ln(1/[H+])
Let us change the base to 10 by multiplying with 2.303.
E = E0 – (2.303RT/nF)log10(1/[H+])
By definition we have – log10[H+] = log10(1/[H+]) = pH
And given that 2.303RT/F = 0.059 V at 298 K,
Using these values we get,
E = E0 – (0.0591/n)pH
We have,
E0 = 0.269 V (calculated)
pH = 2
n = 2 Number of electrons transferred.
E = 0.269 – (0.059/2) x 2
E = 0.269 – 0.059 ……………(2 got cancelled)
E = 0.210 V
EMF of the cell at pH 2 will be 0.210 V.
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