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Electrochemistry question

2022 · 25 Jun · Shift 2 · Q7
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Electrochemistry question

2022 · 25 Jun · Shift 2 · Q7

JEE MainChemistryElectrochemistryMCQ+4 / −1
The correct order of reduction potentials of the following pairs is A. Cl2Cl_2Cl2​/Cl−Cl^-Cl− B. I2I_2I2​/I−I^-I− C. Ag+Ag^+Ag+/Ag D. Na+Na^+Na+/Na E. Li+Li^+Li+/Li Choose the correct answer from the options given below.
  1. A
    A > C > B > D > E
  2. B
    A > B > C > D > E
  3. C
    A > C > B > E > D
  4. D
    A > B > C > E > D
View written solutionFree

Correct answer: A

  1. Write the standard reduction potentials (approximate values):

    • Cl2+2e−→2Cl−Cl_2 + 2e^- \rightarrow 2Cl^-Cl2​+2e−→2Cl−, E∘=+1.36 VE^\circ = +1.36\,\text{V}E∘=+1.36V

    • I2+2e−→2I−I_2 + 2e^- \rightarrow 2I^-I2​+2e−→2I−, E∘=+0.54 VE^\circ = +0.54\,\text{V}E∘=+0.54V

    • Ag++e−→AgAg^+ + e^- \rightarrow AgAg++e−→Ag, E∘=+0.80 VE^\circ = +0.80\,\text{V}E∘=+0.80V

    • Na++e−→NaNa^+ + e^- \rightarrow NaNa++e−→Na, E∘=−2.71 VE^\circ = -2.71\,\text{V}E∘=−2.71V

    • Li++e−→LiLi^+ + e^- \rightarrow LiLi++e−→Li, E∘=−3.04 VE^\circ = -3.04\,\text{V}E∘=−3.04V

  2. Arrange them in decreasing order of reduction potential:

    Compare the values: +1.36>+0.80>+0.54>−2.71>−3.04+1.36 > +0.80 > +0.54 > -2.71 > -3.04+1.36>+0.80>+0.54>−2.71>−3.04

    Therefore, Cl2/Cl−>Ag+/Ag>I2/I−>Na+/Na>Li+/LiCl_2/Cl^- > Ag^+/Ag > I_2/I^- > Na^+/Na > Li^+/LiCl2​/Cl−>Ag+/Ag>I2​/I−>Na+/Na>Li+/Li

  3. Match with the given labels:

    • A =Cl2/Cl−= Cl_2/Cl^-=Cl2​/Cl−
    • B =I2/I−= I_2/I^-=I2​/I−
    • C =Ag+/Ag= Ag^+/Ag=Ag+/Ag
    • D =Na+/Na= Na^+/Na=Na+/Na
    • E =Li+/Li= Li^+/Li=Li+/Li

    So the order is: A>C>B>D>EA > C > B > D > EA>C>B>D>E

  4. Check options:

    • Option A: A>C>B>D>EA > C > B > D > EA>C>B>D>E ✅
    • Option B: A>B>C>D>EA > B > C > D > EA>B>C>D>E ❌
    • Option C: A>C>B>E>DA > C > B > E > DA>C>B>E>D ❌
    • Option D: A>B>C>E>DA > B > C > E > DA>B>C>E>D ❌

Hence, the correct option is A.

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