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Redox Reactions question

2023 · 13 Apr · Shift 2 · Q15
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Redox Reactions question

2023 · 13 Apr · Shift 2 · Q15

JEE MainChemistryRedox ReactionsNumerical+4 / −1
See the following chemical reaction: Cr2O72−+XH++6 Fe2+→YCr3++6 Fe3++ZH2O\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}+\mathrm{XH}^{+}+6 \mathrm{~F}_{e}^{2+} \rightarrow \mathrm{YCr}^{3+}+6 \mathrm{~F}_{e}^{3+}+\mathrm{Z} \mathrm{H}_{2} \mathrm{O}Cr2​O72−​+XH++6 Fe2+​→YCr3++6 Fe3+​+ZH2​O The sum of X,Y\mathrm{X}, \mathrm{Y}X,Y and Z\mathrm{Z}Z is ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 23

  1. We need to balance the ionic redox reaction:

Cr2O72−+X H++6 Fe2+→Y Cr3++6 Fe3++Z H2O\mathrm{Cr_2O_7^{2-}} + X\,\mathrm{H^+} + 6\,\mathrm{Fe^{2+}} \rightarrow Y\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + Z\,\mathrm{H_2O}Cr2​O72−​+XH++6Fe2+→YCr3++6Fe3++ZH2​O

  1. Use the standard dichromate reduction half-reaction in acidic medium:

Cr2O72−+14 H++6e−→2 Cr3++7 H2O\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^+} + 6e^- \rightarrow 2\,\mathrm{Cr^{3+}} + 7\,\mathrm{H_2O}Cr2​O72−​+14H++6e−→2Cr3++7H2​O

  1. Iron(II) is oxidized as:

Fe2+→Fe3++e−\mathrm{Fe^{2+}} \rightarrow \mathrm{Fe^{3+}} + e^-Fe2+→Fe3++e−

Since there are 6 Fe2+6\,\mathrm{Fe^{2+}}6Fe2+, total electrons released are 666, which exactly match the 6e−6e^-6e− required in the dichromate half-reaction.

  1. Therefore, the balanced overall reaction is:

Cr2O72−+14 H++6 Fe2+→2 Cr3++6 Fe3++7 H2O\mathrm{Cr_2O_7^{2-}} + 14\,\mathrm{H^+} + 6\,\mathrm{Fe^{2+}} \rightarrow 2\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + 7\,\mathrm{H_2O}Cr2​O72−​+14H++6Fe2+→2Cr3++6Fe3++7H2​O

  1. Compare with the given form:

Cr2O72−+X H++6 Fe2+→Y Cr3++6 Fe3++Z H2O\mathrm{Cr_2O_7^{2-}} + X\,\mathrm{H^+} + 6\,\mathrm{Fe^{2+}} \rightarrow Y\,\mathrm{Cr^{3+}} + 6\,\mathrm{Fe^{3+}} + Z\,\mathrm{H_2O}Cr2​O72−​+XH++6Fe2+→YCr3++6Fe3++ZH2​O

So,

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

  1. Hence,

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

Therefore, the required integer is:

23\boxed{23}23​

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