Chemistry · Redox Reactions

JEE Main 2026 — 2 April, Morning Shift — Question 46

An electrochemical cell is constructed using half cells in the direction of spontaneous change Fe(OH)2( s)+2e−→Fe(s)+2OH(aq)E0=−0.88 V\mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{e}^{-} \rightarrow \mathrm{Fe}(\mathrm{s})+2 \mathrm{OH}(\mathrm{aq}) \quad \mathrm{E}^{0}=-0.88 \mathrm{~V} and AgBr⁡(s)+e−→Ag⁡(s)+Br⁡−(aq)E0=+0.07 V\operatorname{AgBr}(\mathrm{s})+\mathrm{e}^{-} \rightarrow \operatorname{Ag}(\mathrm{s})+\operatorname{Br}^{-}(\mathrm{aq}) \mathrm{E}^{0}=+0.07 \mathrm{~V} Which of the following option is correct?

  1. Option A:

    Overall reaction \mathrm{Fe}(\mathrm{s})+2 \mathrm{OH}^{1}(\mathrm{aq})+2 \mathrm{AgBr}(\mathrm{s})$$\rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{~s})+2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{Br}^{-}(\mathrm{aq})

    Correct
  2. Option B:

    Ecell 0=−0.95 V\mathrm{E}_{\text {cell }}^{0}=-0.95 \mathrm{~V}

  3. Option C:

    Fe is reduced in the electrochemical cell

  4. Option D:

    Ecell 0\mathrm{E}_{\text {cell }}^{0} is an extensive property

Answer: A

Step-by-step solution

Overall reaction Fe(s)+2OH−(aq)+2AgBr(s)⇌Fe(OH)2( s)+2Ag(s)+2Br−(aq)\mathrm{Fe}(\mathrm{s})+2 \mathrm{OH}^{-}(\mathrm{aq})+2 \mathrm{AgBr}(\mathrm{s}) \rightleftharpoons \mathrm{Fe}(\mathrm{OH})_{2}(\mathrm{~s})+ 2 \mathrm{Ag}(\mathrm{s})+2 \mathrm{Br}^{-}(\mathrm{aq}) Ecell o=ER( cathode )o−ER( anode )o\mathrm{E}_{\text {cell }}^{o}=\mathrm{E}_{\mathrm{R}(\text { cathode })}^{o}-\mathrm{E}_{\mathrm{R}(\text { anode })}^{o} =0.07−(−0.88)=0.95 V=0.07-(-0.88)=0.95 \mathrm{~V}

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Subject
Chemistry
Chapter
Redox Reactions
Topic
n-Factor, Redox Titrations, Self Indicator & Miscellaneous Cases
An electrochemical cell is constructed using half cells in the… | JEE Main 2026 PYQ with Solution · DhiX AI