JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
For the disproportionation reaction \(2 Cu ^{+}( aq ) \rightleftharpoons Cu ( s )+ Cu ^{2+}( aq )\) at \(298\, K\) In \(K\) (where \(K\) is the equilibrium constant) is....... \(\times 10^{-1}\) Given \(\left( E _{ Cu ^{2+} / Cu ^{+}}^{0}=0.16 V \right.\) \(E _{ Cu ^{+} / Cu }^{0}=0.52 V\)\(\left.\frac{ RT }{ F }=0.025\right)\)
- A \(140\)
- B \(144\)
- C \(150\)
- D \(156\)
Answer & Solution
Correct Answer
(B) \(144\)
Step-by-step Solution
Detailed explanation
\(Cu ^{+} \longrightarrow Cu + e ^{-}\) \(Cu ^{+}+ e ^{-} \longrightarrow Cu ( s )\) __________________________________ \(2 Cu ^{+} \longrightarrow Cu ^{2+}+ Cu\) \(E_{c e l l}^{o}=E_{C u^{+} / C u}^{0}-E_{C u^{2+} / C u^{+}}^{0}\) \(=0.52-0.16\) \(=0.36 V\) At equilibrium…
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