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JEE Main 201911 JanuaryMorning Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2019 · 11 January · Morning Shift · Chemistry · Electrochemistry

For the cell Zn(s)|Zn^ 2+ (aq) \| M^ x+ (aq)| M(s), different half...

For the cell Zn(s)∣Zn2+(aq)∥Mx+(aq)∣M(s),\mathrm{Zn}(\mathrm{s})\left|\mathrm{Zn}^{2+}(\mathrm{aq}) \| \mathrm{M}^{\mathrm{x}+}(\mathrm{aq})\right| \mathrm{M}(\mathrm{s}), different half cells and their standard electrode potentials are given below: If \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}^{\circ}=-0.76 \mathrm{ V}, which cathode will give a maximum value of E_{cell 0\text{cell }^{0}perelectrontransferred?per electron transferred?

Options

  1. AAg+/Ag\mathrm{Ag}^{+} / \mathrm{Ag}
  2. BFe3+/Fe2+\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}
  3. CAu3+/Au\mathrm{Au}^{3+} / \mathrm{Au}
  4. DFe2+/Fe\mathrm{Fe}^{2+} / \mathrm{Fe}

Correct answer

A. Ag+/Ag\mathrm{Ag}^{+} / \mathrm{Ag}

Step-by-step solution

Zn+2Ag+⟶2Ag+Zn2+(n=2e−)=0.80−(−0.76)=1.56 V\mathrm{Zn}+2 \mathrm{Ag}^{+} \longrightarrow 2 \mathrm{Ag}+\mathrm{Zn}^{2+}\left(n=2 e^{-}\right) =0.80-(-0.76)=1.56 \mathrm{~V} for 2e∴Ecell ∘2 e \therefore \mathrm{E}_{\text{cell }}^{\circ} for 1e−=1.562=0.78 V1 e^{-}=\frac{1.56}{2}=0.78 \mathrm{~V}
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