A solution of KMnO_4 is standardized using 1.020 g of FAS (M = 392.14 g/mol). If

Question

A solution of KMnO_4 is standardized using 1.020 g of FAS (M = 392.14 g/mol). If the titration requires 25.62 mL of permanganate solution to reach the endpoint, what is the concentration of the permanganate solution? 5 Fe^2+ MnO^-_4 + 8 H^+ rightarrow 5 Fe^3+ + Mn^2+ + 4 H_2O A 2.358 g sample containing iron(II) is titrated using the standardization solution from question If the titration requires 33.55 mL of titrant to reach the endpoint, what is the mass percent of iron in the sample? Balance the following redox reaction in acidic solution: PbO_2(s) + I^-(aq) rightarrow Pb^2+(aq) + I_2(s) Look up the values for SRP's and determine the standard cell potential for the following cell: Zn(s) | Zn^2+ || Cu^2+ | Cu(s) Then determine the cell potential when [Zn^2+]= 0.00101 M and [Cu^2+] = 0.550 M. Repeat #8 for the reaction in question #7. Determine the cell potential when [I^-] = 0.223 M [Pb^2+] = 0.0505 M.

Explanation / Answer

5)

5Fe2+ + MnO4- + 8H+ -----> 5Fe3+ + Mn2+ + 4H2O

5mol of Fe2+ react with 1mol of MnO4-

Mass of FAS = 1.020g

Molar mass of FAS = 392.14g

No of mol of FAS = 1.020g/392.14g = 0.00260

FAS contain 1 Fe2+,

So, No of mol of Fe2+ = 0.00260

0.00260 mol of Fe2+ react with (0.00260/5)= 0.00052 mole of MnO4-

Volume of KMnO4 = 25.62

Concentration of MnO4- = (0.00052/25.62)×1000=0.0203M

[ MnO4- ] = [ KMnO4 ]

Therefore , Concentration of KMnO4 = 0.0203M

6) volume KMnO4 Consumed = 33.55ml

[ KMnO4] = 0.0203M

No of moles of KMnO4 =( 0.0203/1000)× 33.55= 0.000681

No of mole of Fe2+ reacted = 5 × 0.000681= 0.003405

Molar mass of Fe = 55.85g

Mass of Fe = 0.003405 × 55.85g = 0.1902g

% of Fe = (0.1902g/2.358g)× 100 = 8.07%

7) PbO2(s) + 2I-(aq) + 4H+ -------> Pb2+(aq) + I2(s) + 2H2O(l)

8) Zn(s ) ------> Zn2+(aq) + 2e- Anode E°red = -0.763V

Cu2+(aq) + 2e -------> Cu(s) Cathode E°red = 0.337V

E°(cell) = E°red,cathode - E°red,anode

= 0.337 - (-0.763)

= 1.10V

Nernst equation is

E(cell) = E°(cell) - (0.0592/n)log([Ox]/[Red])

= 1.1 V - (0.0592/2)log(0.0010/0.550)

= 1.18V

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