JEE Advanced 2024 Paper 1 · Chemistry · Redox Reactions and Titrations · medium

A 25.0 mL solution contains 0.020 M FeC2O4 and 0.010 M Fe2(C2O4)3. In acidic medium it is titrated with 0.020 M KMnO4 (which is reduced to Mn^2+).…

Numerical answer

25.0\ \mathrm{mL}

solution contains

0.020\ \mathrm{M}

\mathrm{FeC_2O_4}

and

0.010\ \mathrm{M}

\mathrm{Fe_2(C_2O_4)_3}

. In acidic medium it is titrated with

0.020\ \mathrm{M}

\mathrm{KMnO_4}

(which is reduced to

\mathrm{Mn^{2+}}

). The volume of

\mathrm{KMnO_4}

solution required, in

\mathrm{mL}

, is

Show answer and solution

Answer: 30

Only species that can be oxidised consume permanganate: in

\mathrm{FeC_2O_4}

both

\mathrm{Fe^{2+}\to Fe^{3+}}

(1 e

^-

) and

\mathrm{C_2O_4^{2-}\to 2CO_2}

(2 e

^-

) occur, i.e. 3 e

^-

per formula unit. In

\mathrm{Fe_2(C_2O_4)_3}

iron is already

\mathrm{Fe^{3+}}

, but three oxalate ions each give 2 e

^-

, i.e. 6 e

^-

per formula unit. Amounts in 25.0 mL:

\mathrm{FeC_2O_4}=0.500

mmol, giving

1.50

mmol e

^-

;

\mathrm{Fe_2(C_2O_4)_3}=0.250

mmol, giving

1.50

mmol e

^-

. Total

3.00

mmol e

^-

. Each

\mathrm{MnO_4^-}

accepts 5 e

^-

, so

\mathrm{MnO_4^-}

needed

=0.600

mmol, and

V=0.600/0.020=30

mL.

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