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Redox Reactions question

2020 · 5 Sep · Shift 2 · Q1
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  5. /2020 · 5 Sep · Shift 2 · Q1

Redox Reactions question

2020 · 5 Sep · Shift 2 · Q1

JEE MainChemistryRedox ReactionsNumerical+4 / −1
The volume, in mL, of 0.02 M K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ solution required to react with 0.288 g of ferrous oxalate in acidic medium is ‾\underline{\hspace{2cm}}​. (Molar mass of Fe = 56 g mol–1)
Numerical answer
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Correct answer: 50

  1. Identify the formula of ferrous oxalate

Ferrous oxalate is FeC2O4\mathrm{FeC_2O_4}FeC2​O4​.

Its molar mass is: 56+(2×12)+(4×16)=56+24+64=144 g mol−156 + (2\times 12) + (4\times 16) = 56 + 24 + 64 = 144\ \text{g mol}^{-1}56+(2×12)+(4×16)=56+24+64=144 g mol−1

  1. Calculate moles of ferrous oxalate

Given mass =0.288 g= 0.288\ \text{g}=0.288 g

n(FeC2O4)=0.288144=0.002 moln(\mathrm{FeC_2O_4}) = \frac{0.288}{144} = 0.002\ \text{mol}n(FeC2​O4​)=1440.288​=0.002 mol

So, we have:

  • 0.0020.0020.002 mol of Fe2+\mathrm{Fe^{2+}}Fe2+
  • 0.0020.0020.002 mol of C2O42−\mathrm{C_2O_4^{2-}}C2​O42−​
  1. Determine total electrons lost in acidic medium

In acidic medium:

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

So 111 mol Fe2+\mathrm{Fe^{2+}}Fe2+ loses 111 mol electron. Hence, 0.0020.0020.002 mol Fe2+\mathrm{Fe^{2+}}Fe2+ loses: 0.002 mol e−0.002\ \text{mol e}^-0.002 mol e−

  • Oxalate oxidizes as: C2O42−→2CO2+2e−\mathrm{C_2O_4^{2-} \to 2CO_2 + 2e^-}C2​O42−​→2CO2​+2e−

So 111 mol C2O42−\mathrm{C_2O_4^{2-}}C2​O42−​ loses 222 mol electrons. Hence, 0.0020.0020.002 mol C2O42−\mathrm{C_2O_4^{2-}}C2​O42−​ loses: 0.004 mol e−0.004\ \text{mol e}^-0.004 mol e−

Therefore total electrons lost: 0.002+0.004=0.006 mol e−0.002 + 0.004 = 0.006\ \text{mol e}^-0.002+0.004=0.006 mol e−

  1. Electrons accepted by dichromate

In acidic medium: Cr2O72−+14H++6e−→2Cr3++7H2O\mathrm{Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O}Cr2​O72−​+14H++6e−→2Cr3++7H2​O

Thus, 111 mol K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​ accepts 666 mol electrons.

Required moles of K2Cr2O7\mathrm{K_2Cr_2O_7}K2​Cr2​O7​: n=0.0066=0.001 moln = \frac{0.006}{6} = 0.001\ \text{mol}n=60.006​=0.001 mol

  1. Calculate volume of 0.02 M0.02\,M0.02M K2Cr2O7K_2Cr_2O_7K2​Cr2​O7​ solution

Using M=nV⇒V=nMM = \frac{n}{V}\quad \Rightarrow \quad V = \frac{n}{M}M=Vn​⇒V=Mn​

V=0.0010.02=0.05 LV = \frac{0.001}{0.02} = 0.05\ \text{L}V=0.020.001​=0.05 L

Convert to mL: 0.05 L=50 mL0.05\ \text{L} = 50\ \text{mL}0.05 L=50 mL

  1. Final Answer

The required volume is: 50 mL\boxed{50\ \text{mL}}50 mL​

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