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JEE Main 202313 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
13 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Electrochemistry

JEE Main 2023 · 13 April · Evening Shift · Chemistry · Electrochemistry

At 298K , the standard reduction potential for Cu ^ 2 + / Cu...

At 298K298K, the standard reduction potential for Cu2+/Cu{Cu}^{2 +} / Cu electrode is 0.34V0 . 34V. Given : KspCu(OH)2=1×10−20{K}_{sp}Cu ( OH{)}_{2} = 1 \times {10}^{- 20} Take 2.303RTF=0.059V\dfrac{2 . 303RT}{F} = 0 . 059V The reduction potential at pH=14pH = 14 for the above couple is (−)x×10−2V( - ) x \times {10}^{- 2}V. The value of x is

Correct answer

0

Step-by-step solution

For the reaction:
Cu2+(aq)+2e−→Cu(s){Cu}^{2 +}\left(aq\right) + 2{e}^{-} \to Cu\left(s\right)

The reduction potential is

E=Eο−RT2Fln1[Cu2+]E = {E}^{ο} - \dfrac{RT}{2F}ln\dfrac{1}{\left[{Cu}^{2 +}\right]}

From the given data pH=14pH = 14 and

Ksp=(Cu(OH)2)=1.0×10−20{K}_{sp} = \left(Cu{\left(OH\right)}_{2}\right) = 1 .0 \times {10}^{- 20}, we get

\left[{H}^{+}\right] = {10}^{- 14}M, \left[{OH}^{-}\right] = \dfrac{{K}_{W}}{\left[{H}^{+}\right]} = \dfrac{{10}^{- 14}{M}^{2}}{{10}^{- 14}M} = 1M<br/><br/>[Cu2+]=Ksp[OH−]2=1.0×10−20M21M=1.0×10−20M<br /><br /> \left[{Cu}^{2 +}\right] = \dfrac{{K}_{sp}}{{\left[{OH}^{-}\right]}^{2}} = \dfrac{1 .0 \times {10}^{- 20}{M}^{2}}{1M} = 1 .0 \times {10}^{- 20}M

E=0.34V−(0.059V2)log11.0×10−20=0.34V−0.059×20V2E=0.34−0.59=−0.25VE =0.34 V - \left(\dfrac{0 .059 V}{2}\right)log\dfrac{1}{1 .0 \times {10}^{- 20}} = 0 .34 V - \dfrac{0 .059 \times 20V}{2}E = 0 . 34 - 0 . 59 = - 0 . 25V
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