Equal volumes of 1 M each of KMnO4 and K2Cr2O7 are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be :
Correct answer: C. More with K2Cr2O7
- A. More with KMnO4
- B. Equal with both oxidizing agents
- C. More with K2Cr2O7
- D. Cannot be determined
Explanation
Both KMnO4 and K2Cr2O7 are strong oxidizing agents that can oxidize Fe(II) to Fe(III) in acidic medium. The n-factor of KMnO4 in acidic medium is 5, while for K2Cr2O7, it is 6. This means that for the same number of moles, K2Cr2O7 can transfer more electrons and thus oxidize more moles of Fe(II). Therefore, K2Cr2O7 will oxidize more Fe(II) compared to KMnO4. Option A is incorrect because KMnO4 has a lower n-factor. Option B is incorrect because the n-factors are different, leading to different amounts of Fe(II) oxidized. Option D is incorrect because the comparison of n-factors provides a clear determination.
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? [email protected]
About Electrochemistry
Electrochemistry describes electron transfer through oxidation and reduction, identifies oxidizing and reducing agents, and applies oxidation numbers to balance redox equations. It also covers electrode potential and the standard hydrogen electrode, which provides the reference for comparing the reduction tendencies of other electrodes.
Practise Electrochemistry
699 free Electrochemistry MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Chemistry questions like this
Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Electrochemistry questions
A 10.0 g sample of a mixture of calcium chloride and sodium chloride is treated with Na2CO3 to precipitate the calcium as calcium carbonate. This CaCO3 is heated to convert all the calcium to CaO and the final mass of CaO is 1.62 gms. The % by mass of CaCl2 in the original mixture is:
Find the weight of KOH in its 50 milli equivalents:
10-2 mole of Fe3O4 is treated with excess KI solution in presence of dilute H2SO4, the products are Fe2+ and I2(g). What volume of 0.1 (M) Na2S2O3 will be needed to reduce the liberated I2(g)?
How many millilitre of 0.5 M H2SO4 are needed to dissolve 0.5 g of Cu(II) carbonate?
O.N. of V in Rb4 Na [HV10O28] is:
When an alkaline solution of K2CrO4 is treated with 3% H2O2 solution, red brown paramagnetic peroxochromate is obtained as per following equation: 2K2CrO4 + 7H2O2 + 2KOH → 2K3CrO8 + 8H2O The equivalent weight of K2CrO4 for above transformation must be (assuming M is the molar mass of K2CrO4):