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NVAEasyJEE 2023Oxidation Number & Redox Reactions

JEE Chemistry 2023 Question with Solution

See the following chemical reaction: Cr2O72−+XH++6Fe2+→YCr3++6Fe3++ZH2OCr_2O_7^{2-} + XH^+ + 6Fe^{2+} \rightarrow YCr^{3+} + 6Fe^{3+} + ZH_2O The sum of XX, YY and ZZ is _____ .

Answer

Correct answer:23

Step-by-step solution

Standard Method

Given: The reaction is

Cr2O72−+XH++6Fe2+→YCr3++6Fe3++ZH2OCr_2O_7^{2-} + XH^+ + 6Fe^{2+} \rightarrow YCr^{3+} + 6Fe^{3+} + ZH_2O

Find: The value of X+Y+ZX + Y + Z.

Balance the chemical equation:

Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O

From the balanced equation, the coefficients are:

X=14,Y=2,Z=7X = 14, \quad Y = 2, \quad Z = 7

Therefore,

X+Y+Z=14+2+7=23X + Y + Z = 14 + 2 + 7 = 23

So, the required numerical value is 2323.

Redox Balancing Tip

Given: A dichromate ion is reduced in acidic medium while Fe2+Fe^{2+} is oxidized to Fe3+Fe^{3+}. Find: The sum of the missing coefficients.

Use the balancing idea stated in the solution: first balance elements other than oxygen and hydrogen, then balance oxygen using water, and hydrogen using H+H^+. This gives the balanced equation

Cr2O72−+14H++6Fe2+→2Cr3++6Fe3++7H2OCr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O

Hence X=14X = 14, Y=2Y = 2, and Z=7Z = 7, so the sum is 2323.

Common mistakes

  • A common mistake is to balance only atoms and ignore charge. In redox equations, charge balance is essential. After balancing atoms, verify the ionic charges on both sides before finalizing the coefficients.

  • Students often place the wrong coefficient on H+H^+ or H2OH_2O. In acidic medium, oxygen is balanced using water and hydrogen is then balanced using H+H^+, not the other way around.

  • Another mistake is to read YCr3+YCr^{3+} as a different chromium species instead of a coefficient multiplying Cr3+Cr^{3+}. Here, YY is the stoichiometric coefficient of Cr3+Cr^{3+}.

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