Question

Ratio of mols of Fe(II) oxidised by equal volumes of equilmolar KMnO4and K2Cr2O7 solutions in acidic medium will be

Solution

Correct option is

5 : 6

 

Eq. mass of Fe(II) = molar mass

Eq. mass of KMnO4 = molar mass/5

Eq. mass K2Cr2O7 = molar mass/6

Eqiv. of KMnO4 in VL of  M =5xV =mol of Fe(II)

Eqiv. of K2Cr2O7 in VL of  M = 6xV =mol of Fe(II) 

Hence, ratio of moles of Fe(II) oxidized by KMnO4 and K2Cr2O7 = 5 : 6

SIMILAR QUESTIONS

Q1

1 mol of N2 H4 loses 10 mol of electrons to form a new compound X. Assuming that all the nitrogen appears in the new compound and no change in the oxidation state of hydrogen, the oxidation number in X is

Q2

Which of the following is the strongest reducing agent?

Q3

In which of the following compounds the oxidation number of carbon is not zero?

Q4

When ethane is  burnt in presence of excess of oxygen, the oxidation number of carbon changes by

Q5

In a reaction, 3mol of electrons are gained by 1 mol of HNO3. Assuming no change in oxidation state of hydrogen the possible product obtained due to reduction will be

Q6

The oxidation numbers of most electronegative element in the products of hydrolysis of peroxydisulphuric acid are

Q7

 

For the redox reaction  the correct coefficients of the reactants for the balanced equation are

                                            

Q8

The equivalent  mass of Na2CrO4 used as oxidant in acidic medium is (Cr = 52)

Q9

The equivalent masses of Na2CO3 when it is titrated with HCl using phenolphthalein and methyl orange indicators in separate experiments are

Q10

Which of the following can be reduced by SnCl2 solution