The reaction of MnO4^? with oxalic acid (H2C2O4) in acidic solution, yielding Mn
ID: 832249 • Letter: T
Question
The reaction of MnO4^? with oxalic acid (H2C2O4) in acidic solution, yielding Mn^2+ and CO2 gas, is widely used to determine the concentration of permanganate solutions.
A) Write a balanced net ionic equation for the reaction.
B) Use the data in Appendix D in the textbook to calculate E? for the reaction.
C) and D) Show that the reaction goes to completion by calculating the values of ?G? and pK=?log(K) at 25 ?C
E) A 1.325g sample of sodium oxalate (Na2C2O4) is dissolved in dilute H2SO4 and then titrated with a KMnO4 solution. If 30.95mL of the KMnO4 solution is required to reach the equivalence point, what is the molarity of the KMnO4 solution? Express your answer using four significant figures.
Explanation / Answer
(A) Oxidation: 5 H2C2O4 => 10 CO2 + 10 H+ + 10 e-, Eo(ox) = 0.49 V
Reduction: 2 MnO4- + 16 H+ + 10 e? => 2 Mn2+ + 8 H2O, Eo(red) = 1.51 V
Add Oxidation + Reduction and cancel common terms to get overall reaction:
2 MnO4-(aq) + 5 H2C2O4(aq) + 6 H+(aq) => 2 Mn2+(aq) + 10 CO2(g) + 8 H2O(l)
(B) Eo = Eo(ox) + Eo(red)
= 0.49 + 1.51 = 2.00 V
(C) Moles of electrons transferred = n = 10
Faraday constant F = 96485 C/mol
Temperature T = 25 deg C = 298.15 K
Molar gas constant R = 8.314 J/mol.K
Delta Go = -nFEo
= -10 x 96485 x 2.00
= -1929700 J = -1929.7 kJ (approximately -1.93 x 10^3 kJ)
K = exp(-Delta Go/RT)
= exp(1929700/(8.314 x 298.15))
= 1.224 x 10^338
pK = -log K = -log(1.224 x 10^338) = -338
Since Delta Go is negative => reaction is spontaneous and goes to completion under standard state conditions.
(D) Moles of H2C2O4 = moles of Na2C2O4 = mass/molar mass of Na2C2O4
= 1.325/134.0 = 0.00988806 mol
Moles of MnO4- = 2/5 x moles of H2C2O4
= 2/5 x 0.00988806 = 0.0039552 mol
Molarity of KMnO4 = moles/volume of MnO4-
= 0.0039552/0.03095 = 0.1278 M
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