Virat Batch 2027
Champion Batch 2028
Divine JEE 1-on-1
At 298 K298 \mathrm{~K}298 K, the standard reduction potential for Cu2+/Cu\mathrm{Cu}^{2+} / \mathrm{Cu}Cu2+/Cu electrode is 0.34 V0.34 \mathrm{~V}0.34 V.
Given : KspCu(OH)2=1×10−20\mathrm{K}_{\mathrm{sp}} \mathrm{Cu}(\mathrm{OH})_{2}=1 \times 10^{-20}KspCu(OH)2=1×10−20
Take 2.303RTF=0.059 V\frac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059 \mathrm{~V}F2.303RT=0.059 V
The reduction potential at pH=14\mathrm{pH}=14pH=14 for the above couple is (−)x×10−2 V(-) x \times 10^{-2} \mathrm{~V}(−)x×10−2 V. The value of xxx is ___________
Cu(OH)2( s) ⇌ Cu2+(aq)+2 OH−(aq) \mathrm{Cu}(\mathrm{OH})_2(\mathrm{~s}) \rightleftharpoons \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq}) \\ Cu(OH)2( s) ⇌ Cu2+(aq)+2 OH−(aq)
Ksp=[Cu2+][OH−]2 \mathrm{Ksp}=\left[\mathrm{Cu}^{2+}\right]\left[\mathrm{OH}^{-}\right]^2 \\ Ksp=[Cu2+][OH−]2
pH=14 ; pOH=0 ;[OH−]=1 M \mathrm{pH}=14 ; \mathrm{pOH}=0 ;\left[\mathrm{OH}^{-}\right]=1 \mathrm{M} \\ pH=14 ; pOH=0 ;[OH−]=1 M
[Cu2+]=Ksp[1]2=10−20 M {\left[\mathrm{Cu}^{2+}\right]=\frac{\mathrm{Ksp}}{[1]^2}=10^{-20} \mathrm{M}} \\ [Cu2+]=[1]2Ksp=10−20 M
Cu2+(aq)+2 e− → Cu( s) \mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{~s}) \\ Cu2+(aq)+2 e− → Cu( s)
E=E∘−0.0592 log 10 1[Cu2+] \mathrm{E}=\mathrm{E}^{\circ}-\frac{0.059}{2} \log _{10} \frac{1}{\left[\mathrm{Cu}^{2+}\right]} \\ E=E∘−20.059 log 10 [Cu2+]1
=0.34−0.0592 log 10 110−20 =0.34-\frac{0.059}{2} \log _{10} \frac{1}{10^{-20}} \\ =0.34−20.059 log 10 10−201
=−0.25=−25 × 10−2 =-0.25=-25 \times 10^{-2} =−0.25=−25 × 10−2
Select an option to instantly check whether it is correct or wrong.
Learn on the go with our mobile app. Access courses, live classes, and study materials anytime, anywhere.
Empowering students to achieve their IIT dreams through personalized learning and expert guidance.
Copyright © 2026 Divine JEE. All Rights Reserved.