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JEE Advanced 2018Paper 1
Exam
JEE Advanced
Year
2018
Paper
Paper 1
Subject
Chemistry
Chapter
Electrochemistry

JEE Advanced 2018 · Paper 1 · Chemistry · Electrochemistry

For the electrochemical cell, Mg(s)| Mg ^ 2+ (aq,1M)| | Cu ^ 2+...

For the electrochemical cell, \text{Mg(s)}\left|{\text{Mg}}^{\text{2+}} \left(\text{aq,1}\text{M}\right)\right|\left|{\text{Cu}}^{\text{2+}} \left(\text{aq,1}\text{M}\right)\right|\text{Cu}\left(\text{s}\right)thethestandard emf of the cell is \text{2}\text{.70 V}atat\text{300 K}\text{.} Whentheconcentrationothe concentration of{\text{Mg}}^{\text{2+}}ischangedtois changed to\text{x M,}thecellpotentialchangestothe cell potential changes to\text{2}\text{.67 V}atat\text{300 K}\text{.}TheThevalue of\text{x}is ________.<br /> <br /> (given,\dfrac{\text{F}}{\text{R}}{\text{=11500 K V}}^{- \text{1}} \text{,}where \text{F} isthetheFaraday constant and \text{R} is the gas constants,\text{ln}\left(\text{10}\right)\text{=}\text{2}\text{.30} )

Correct answer

0

Step-by-step solution

Mg(s)+Cu\text{Mg}\left(\text{s}\right) \text{+}{\text{Cu}}^{\text{2+}} \left(\text{aq}\right) \to {\text{Mg}}^{\text{2+}} \left(\text{aq}\right)+\text{+}Cu(s)\text{Cu}\left(\text{s}\right)
{E\text{E}}_{cell\text{cell}}^{o\text{o}} =\text{=}2\text{2}.70, E{\text{.70, E}}_{cell\text{cell}}=\text{=}2\text{2}{.67, Mg\text{.67, Mg}}^{2+\text{2+}} =\text{=}xM,\text{xM,}{\text{Cu}}^{\text{2+}} =\text{=}1\text{1}M<br/>\text{M}<br />{\text{E}}_{\text{cell}} =E\text{=}{\text{E}}_{\text{cell}}^{\text{o}} - \dfrac{RT\text{RT}}{nF\text{nF}} ln\text{ln}x\text{x}<br/>2<br />\text{2}.67\text{.67}=\text{=}2\text{2}.70\text{.70} -\dfrac{\text{RT}}{2F\text{2F}}ln\text{ln}x\text{x}
−0- \text{0}.03\text{.03}=\text{=} -\dfrac{\text{R}\text{\times}\text{300}}2F{\text{2F}} \times\text{\times}ln\text{ln}x<br/>lnx\text{x}<br />\text{ln}\text{x}=\text{=}\dfrac{0\text{0}.03\text{.03}\times\text{\times}2\text{2}}{300\text{300}}\text{\times}\dfrac{\text{F}}{R\text{R}}<br />\text{=}\dfrac{\text{0}\text{.03}\text{\times}\text{2}\text{\times}\text{11500}}{\text{300}\text{\times}\text{1}}
lnx=\text{ln}\text{x}\text{=}2\text{2}.30\text{.30}=\text{=}ln(10)\text{ln}\left(\text{10}\right)
x=10\text{x}\text{=}\text{10}
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