Question

How many moles of $FeC_{2}O_{4}$ are required to react completely with one equivalent of $K_{2}Cr_{2}0_{7}$ in acidic medium ?​

Answer 1

Answer:

1)0.6

2)KMnO4 oxidises both ferrous ions to ferric ions and oxalate ions to carbon dioxide as we can see that the oxidation number ofC atom in oxalate ion(upper pic)is +3,while the oxidation state of Catom in carbon dioxide is +4(lower pic).

3)So, one mole of oxalate ions loose2electrons(one electron for eachC atom)while one the other hand one mole of ferrous ion looses1 electron so one mole of ferrous oxalate molecule looses 3electrons in  total so the required number ofKMnO4should provide those 3electrons as in acidic medium KMnO4 gets reduced to Mn2+ ions so one mole ofKMnO4 gains5 electrons

4)Hence,3/5ie−0.6 mole ofKMnO4is needed to react completely(oxidise completely) one mole of ferrous oxalate.

Answer 2

Explanation:

The kinetics and mechanism of the oxidation of [CrIII(DPA)(IDA)(H2O)]− (DPA = dipicolinate and IDA = iminodiacetate) by periodate in the presence of Mn(II) as a catalyst have been investigated. The rate of the reaction increases with increasing pH, due to the deprotonation equilibria of the complex. Addition of Mn(II) in the concentration range of (2.5–10) × 10−6 mol dm−3 enhanced the reaction rate; the reaction is first order with respect to both [IO4 −] and the Cr complex, and obeys the following rate law: Rate=[CrIII(DPA)(IDA)(H2O)−][MnIII]{(k7+K1k8/[H+])+[IVII]((k9k11/k−9+k11)+(K1k10k12)/(k−10+k12)[H+])}. Catalysis by Mn(II) is believed to be due to initial oxidation of Mn(II) to Mn(III), which acts as the oxidizing agent. It is proposed that electron transfer proceeds through an inner-sphere mechanism via coordination of IO4 − to Cr(III). Thermodynamic activation parameters were calculated using the transition state theory equation.