The amount of manganese in steel is determined by changing it to permanganate io

Question

The amount of manganese in steel is determined by changing it to permanganate ion. The steel is first dissolved in nitric acid, producing Mn+2 ions and NO gas. These Mn+2 ions are then oxidized to the deeply colored MnO4- ions by periodate ion (IO4-) in acidic solution. a) Write the electron configuration of Mn+2. b) Determine the oxidation number of the manganese atom in MnO4-. c) Balance the equation for the reaction between solid manganese and nitric acid. d) Balance the following equation which occurs in acidic media. Mn+2(aq) + IO4-(aq) ? MnO4-(aq) + IO3-(aq) e) If it takes 32.40 mL of 0.7625 M NaIO4 to titrate the manganese ion produced from dissolving 5.56 g of steel in nitric acid, what is the percent manganese in the steel?

Explanation / Answer

a)

Mn=25= 1s2 2s2 2p6 3s2 3p6 4s2 3d5

Mn+2= 23e- = 1s2 2s2 2p6 3s2 3p6 4s0 3d5

b)MnO4-

oxdiation state of Oxygen= -2

x+4(-2) = -1

x= +7

so the oxidation state of Mn in MNO4- = +7

c)Mn(s) + HNO3(aq) ----------------- Mn(NO3)2 + NO + H2O

3 Mn(s) + 8 HNO3(aq) -------------------- 3 Mn(NO3)2 + 2 NO +4 H2O

d)        Mn+2 + IO4 - --------------- MnO4- + IO3-

          +2           +7                      +7         +5

oxidation half reaction

Mn+2 ----------- MnO4-

Mn+2 + 4H2O --------------- MnO4-

Mn+2 + 4H2O ---------------- MnO4- + 8 H+

Mn+2 + 4 H2O --------------- MnO4- + 8 H+ + 5 e-

Reductiion half reaction

IO4- ------------------ IO3-

IO4- --------------- IO3- + H2O

IO4- + 2 H+ ------------ IO3- + H2O

IO4- + 2 H+ + 2e- ------------- IO3- + H2O

[Mn+2 + 4 H2O --------------- MnO4- + 8 H+ + 5 e-]x2

[IO4- + 2 H+ + 2e- ------------- IO3- + H2O]x5

2Mn+2 + 8 H2O ---------------2 MnO4- + 16 H+ + 10 e-

5IO4- + 10 H+ + 10e- -------------5 IO3- +5 H2O

---------------------------------------------------------------------------------------

2 Mn+2 + 5 IO4- + 3H2O -------------- 2 MnO4- + 5 IO3- + 6 H+

---------------------------------------------------------------------------------------------

e)

NaIO4- = 32.40ml of 0.7625M

number o fmoles of IO4- = 0.7625Mx0.03240L= 0.024705 moles

according to equation

5 moles of IO4- ---------- 2 moles of Mn+2

0.024705 moles of IO4- ------- ?

                                        = 2x0.024705/5 = 0.009882 moles of Mn+2

number of moles of Mn+2 = 0.009882 moles

molar mass of Mn+2 = 54.95 gram/mole

mass of 0.009882 moles of Mn+ 2= 0.009882 x54.95 = 0.543 grams.

mass of Mn+2 = 0.543 grams

% of the Mn = 0.543/5.56 x100= 9.77%

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