JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
At \(298\,K\), the standard reduction potential for \(Cu ^{2+} / Cu\) electrode is \(0.34\,V\). Given : \(K _{ sp } Cu ( OH )_2=1 \times 10^{-20}\) \(\operatorname{Take} \frac{2.303 RT }{ F }=0.059 \,V\) The reduction potential at \(pH =14\) for the above couple is \((-) x \times 10^{-2}\,V\). The value of \(x\) is \(........\).
- A \(24\)
- B \(23\)
- C \(22\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
Detailed explanation
\(Cu ( OH )_2( s ) \rightleftharpoons Cu ^{2+}( aq )+2 OH ^{-}( aq )\) \(Ksp =\left[ Cu ^{2+}\right]\left[ OH ^{-}\right]^2\) \(pH =14 ; pOH =0 ;\left[ OH ^{-}\right]=1\,M\) \({\left[ Cu ^{2+}\right]=\frac{ Ksp }{[1]^2}=10^{-20} M }\)…
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