If equal volumes of $1 M$ $K M n O_{4}$ and $1 M$ $K_{2} C r_{2} O_{7}$ solutions are used to oxidise $F e^{2 +}$ in acidic medium, then $F e^{2 +}$ will be oxidised

more by

K_{2} C r_{2} O_{7}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

more by

K M n O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

equal in both the cases

No worries! We‘ve got your back. Try BYJU‘S free classes today!

the data is insufficient to predict the answer

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A more by $K_{2} C r_{2} O_{7}$
The redox half cell reactions are written below:
$F e^{2 +} \to F e^{3 +} + 1 e^{-}$

$M n O_{4}^{⊝} + 8 H^{⨁} + 5 e^{-} \to M n^{2 +} + 4 H_{2} O$

$C r_{2} O_{7}^{2 -} + 14 H^{⨁} + 6 e^{-} \to 2 C r^{3 +} + 7 H_{2} O$

Using mole concept,
$1$ mol $M n O_{4}^{⊝} \equiv 5$ mol $F e^{2 +}$

and $1$ mol $C r_{2} O_{7}^{2 -} \equiv 6$ mol $F e^{2 +}$

So, $K_{2} C r_{2} O_{7}$ will oxidise more $F e^{2 +}$ to $F e^{3 +}$ .

Suggest Corrections

Similar questions

Q. Equal volumes of 1 M each of

K M n O_{4}

and

K_{2} C r_{2} O_{7}

are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be:

Equal volumes of 1 M each of $K M n O_{4}$ and $K_{2} C r_{2} O_{7}$ are used to oxidise Fe(II) solution in acidic medium. The amount of Fe oxidized will be :

If equal volumes of 0.1 M $K M n O_{4}$ and $0.1 M K_{2} C r_{2} O_{7}$ solutions are allowed to oxidise $F e^{2 +}$ to $F e^{3 +}$ in acidic medium, then $F e^{2 +}$ oxidized will be:

Q. Ratio of moles of Fe(||) oxidised by equal volumes of equimolar KMnO₄ and K₂Cr₂O₇ solutions in acidic medium will be:

Q. If equal volumes of

0.1 M K M n O_{4}

and

0.1 M K_{2} C r_{2} O_{7}

solutions are allowed to oxidise

F e^{2 +} t o F e^{3 +}

in acidic medium, then

F e^{2 +}

oxidised will be :

Join BYJU'S Learning Program

If equal volumes of 1M KMnO4 and 1M K2Cr2O7 solutions are used to oxidise Fe2+ in acidic medium, then Fe2+ will be oxidised

The correct option is A more by K2Cr2O7The redox half cell reactions are written below:Fe2+→Fe3++1e−MnO⊝4+8H⨁+5e−→Mn2++4H2OCr2O2−7+14H⨁+6e−→2Cr3++7H2OUsing mole concept,1 mol MnO⊝4≡5 mol Fe2+and 1 mol Cr2O2−7≡6 mol Fe2+So, K2Cr2O7 will oxidise more Fe2+ to Fe3+.

If equal volumes of $1 M$ $K M n O_{4}$ and $1 M$ $K_{2} C r_{2} O_{7}$ solutions are used to oxidise $F e^{2 +}$ in acidic medium, then $F e^{2 +}$ will be oxidised