The equations for the redox reaction are:
#2×["Fe(CN)"_6^"3-" + "e"^"-" → "Fe(CN)"_6^"4-"]#
#1×["AscH"_2 → "Asc" + "2H"^"+" + "2e"^"-"#
#stackrel(————————————————————)("2Fe(CN)"_6^"3-" +"AscH"_2 → "2Fe(CN)"_6^"4-" + "Asc" + "2H"^"+")#
1. Calculate the moles of #"Fe(CN)"_6^"3-"#
#"Moles of Fe(CN)"_6^"3-" = "0.0439" color(red)(cancel(color(black)("L Fe(CN)"_6^"3-"))) ×(1.12 × 10^"-4" color(white)(l)"mol Fe(CN)"_6^"3-")/(1 color(red)(cancel(color(black)("L Fe(CN)"_6^"3-")))) = 4.917 × 10^"-6"color(white)(l) "mol Fe(CN)"_6^"3-"#
2. Calculate the moles of #"AscH"_2#
#"Moles of AscH"_2 = 4.917 × 10^"-6"color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-"))) × ("1 mol AscH"_2)/(2 color(red)(cancel(color(black)("mol Fe(CN)"_6^"3-")))) = 2.458 × 10^"-6"color(white)(l)"mol AscH"_2#
3. Calculate the molarity of the #"AscH"_2#
#"Molarity" = "moles"/"litres" = (2.458 × 10^"-6"color(white)(l) "mol")/("0.0329 L") = 7.47 × 10^"-5"color(white)(l) "mol/L"#