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JEE Main 202229 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
29 July
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2022 · 29 July · Evening Shift · Chemistry · Electrochemistry

For a cell, Cu(s) Cu ^ 2 + (0 . 001M ) | Ag ^ + (0 . 01M)|Ag(s))the...

For a cell, Cu(s)Cu2+(0.001M)∣Ag+(0.01M)∣Ag(s))thecellpotentialisfoundtobeCu\left(s\right) {Cu}^{2 +}\left(0 . 001M ) \left|{Ag}^{+}\left(0 . 01M\right)\right|Ag\left(s\right)\right)the cell potential is found to be0 . 43Vatat298K.Themagnitudeofstandardelectrodepotentialfor. The magnitude of standard electrode potential for{Cu}^{2 +} | Cuis____\times {10}^{- 2}V</p><p>[Given:EAg+/AgΘ=0.80Vand2.303RTF=0.06V]</p><p>\left[Given: {E}_{{Ag}^{+} / Ag}^{Θ} = 0 . 80Vand\dfrac{2 . 303RT}{F} = 0 . 06V\right]

Correct answer

0

Step-by-step solution

At anode Cu→Cu2++2e−Cu \to {Cu}^{2 +} + 2{e}^{-}

At cathode 2Ag++2e−→2Ag2{Ag}^{+} + 2{e}^{-} \to 2Ag

Cell reaction →Cu+2Ag+→Cu2++2Ag\to Cu + 2{Ag}^{+} \to {Cu}^{2 +} + 2Ag

Ecell=Ecell0−0.062log[Cu2+][Ag+]2{E}_{cell} = {E}_{cell}^{0} - \dfrac{0 . 06}{2}log\dfrac{\left[{Cu}^{2 +}\right]}{{\left[{Ag}^{+}\right]}^{2}}

0.43=Ecell0−0.062log(0.001)(0.01)20 . 43 = {E}_{cell}^{0} - \dfrac{0 . 06}{2}log\dfrac{\left(0 . 001\right)}{{\left(0 . 01\right)}^{2}}

Ecell0=0.46{E}_{cell}^{0} = 0 . 46

Ecell0=EAg+/Ag0−ECu2+/Cu0{E}_{cell}^{0} = {E}_{{Ag}^{+} / Ag}^{0} - {E}_{{Cu}^{2 +} / Cu}^{0}

0.46=0.80−ECu2+/Cu00 . 46 = 0 . 80 - {E}_{{Cu}^{2 +} / Cu}^{0}

ECu2+/Cu0=0.34{E}_{{Cu}^{2 +} / Cu}^{0} = 0 . 34 volt

ECu2+/Cu0=34×10−2{E}_{{Cu}^{2 +} / Cu}^{0} = 34 \times {10}^{- 2}

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