6 H+ (aq) + 2 MnO4- (aq) + 5 H2C2O4 (aq) --> 10 CO2 (g) + 2 Mn2+ (aq) + 8 H2O (l

Question

6 H+ (aq) + 2 MnO4- (aq) + 5 H2C2O4 (aq) --> 10 CO2 (g) + 2 Mn2+ (aq) + 8 H2O (l)

(a) What species is oxidized in the reaction? ______________________

(b) What species is reduced in the reaction? ______________________

Suppose a 0.193 g sample of a compound containing oxalate ion is titrated to an endpoint with 23.8 mL of 0.0200 M KMnO4.

(c) What is the mass of C2O42- (MM = 88.0 g/mol) in the sample?

(d) What is the % mass of C2O42- in the sample?

(e) If the theoretical % mass of C2O42- in the compound is 53.8%, what is the % purity of this sample?

(f) What is the oxidation state of Fe in the complex ion, [Fe(C2O4)3]3-?

(g) What is the abbreviated electron configuration of Fe in the complex ion?

(h) The oxalate ion is a medium field ligand stronger than halides but weaker than water. How many unpaired electrons would you predict in a high-spin configuration?

(i) What is the correct name of K3[Fe(C2O4)3]?

Explanation / Answer

a) Carbon get oxidised (C+3 to C+4)

b) Mn get reduced (Mn+7 to Mn+2)

c) The no. of moles of titrant KMnO4 is added, n = 0.23 Lit * 0.20 M = 0.000476 moles. calculated using formula (M=n/V)

According to the balanced equation 2 moles of KMnO4 reacts ---------------> 5 moles of C2O4

So 0.000476 moles --------------------------------> ???

0.00119 moles of C2O4 = 0.00119 mole * 88 gr/mole = 0.1047 gr of C2O4

d) % of mass of C2O4 = (mass of C2O4 / mass of sample)* 100 = (0.1047/0.193)* 100 = 54.2 %

e) % purity = (yielded/ theoritical)* 100 = (54.2/53.8) * 100 = 100.8 %

f) Fe oxidation state is +3, 3 oxalate ions are having -2 charge each and overall charge is -3.

x+ 3(-2) = -3 ==> x=+3.

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