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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2020 · 2 September · Evening Shift · Chemistry · Electrochemistry

For the disproportionation reaction 2 Cu ^ + (aq) ⇌ Cu (s) + Cu ^ 2...

For the disproportionation reaction 2Cu+2{Cu}^{+}\left(aq\right) ⇌ Cu(s)+Cu2+(aq)\left(s\right) + {Cu}^{2 +}\left(aq\right) at 298K,298K,lnKlnK (where KK is the equilibrium constant) is _______×10−1\times {10}^{- 1}

Given :(E°Cu2+/Cu+=0.16VE°Cu+/Cu=0.52VRTF=0.025): \left(E{°}_{{Cu}^{2 +} / {Cu}^{+}} = 0 . 16VE{°}_{{Cu}^{+} / Cu} = 0 . 52V\dfrac{RT}{F} = 0 . 025\right)

Correct answer

0

Step-by-step solution

{E}_{cell}^{o} = {E}_{{Cu}^{+} / Cu}^{°} - {E}_{{Cu}^{2 +} / {Cu}^{+}}^{°}</p><p></p><p>= 0 . 52 - 0 . 16 = 0 . 36V</p><p></p><p>{E}_{cell}^{°} = \dfrac{RT}{nF}{lnK}_{eq}</p><p></p><p>0 . 36 = \dfrac{0 . 025}{1}lnk</p><p>Ink=14.4=144×10−1</p><p>Ink = 14 . 4 = 144 \times {10}^{- 1}

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