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Ionic Equilibrium question

2022 · 27 Jul · Shift 1 · Q21
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Ionic Equilibrium question

2022 · 27 Jul · Shift 1 · Q21

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
In the titration of KMnO4\mathrm{KMnO}_{4}KMnO4​ and oxalic acid in acidic medium, the change in oxidation number of carbon at the end point is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 1

  1. In acidic medium, oxalic acid is oxidized by permanganate.

  2. The relevant oxidation half-reaction for oxalic acid is: H2C2O4→2CO2+2H++2e−\mathrm{H_2C_2O_4 \rightarrow 2CO_2 + 2H^+ + 2e^-}H2​C2​O4​→2CO2​+2H++2e−

  3. Find the oxidation number of carbon in oxalic acid, H2C2O4\mathrm{H_2C_2O_4}H2​C2​O4​. Let oxidation number of each carbon be xxx.

Using: 2(+1)+2x+4(−2)=02(+1) + 2x + 4(-2) = 02(+1)+2x+4(−2)=0 2+2x−8=02 + 2x - 8 = 02+2x−8=0 2x−6=02x - 6 = 02x−6=0 2x=62x = 62x=6 x=+3x = +3x=+3

So, carbon in oxalic acid has oxidation number +3+3+3.

  1. In CO2\mathrm{CO_2}CO2​, oxidation number of carbon is: x+2(−2)=0x + 2(-2) = 0x+2(−2)=0 x−4=0x - 4 = 0x−4=0 x=+4x = +4x=+4

  2. Therefore, during the titration, carbon changes from +3+3+3 to +4+4+4. So the change in oxidation number of each carbon atom is: +4−(+3)=+1+4 - (+3) = +1+4−(+3)=+1

  3. Hence, the required change in oxidation number of carbon is: 1\boxed{1}1​

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