JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The cell potential for \(Zn \left| Zn ^{2+}( aq ) \| Sn ^{ x +}\right| Sn\) is \(0.801\) \(V\) at \(298\,K\). The reaction quotient for the above reaction is \(10^{-2}\). The number of electrons involved in the given electrochemical cell reaction is. .... (Given \(E _{ Zn ^{2+} \mid Zn }^{0}=-0.763\,V , E _{ Sn ^{ x +} \mid Sn }^{0}=+0.008\,V\) and \(\left.\frac{2.303\,RT }{ F }=0.06\,V \right)\)
- A \(8\)
- B \(9\)
- C \(3\)
- D \(2\)
Answer & Solution
Correct Answer
(D) \(2\)
Step-by-step Solution
Detailed explanation
\(E = E ^{0}-\frac{2.303 RT }{ nF } \log Q\) Here, \(E =+0.801\,V , E ^{0}=0.008-(-0.763)\) \(=+0.771\,V\) \(\therefore 0.801=+0.771-\frac{0.06}{ n } \log 10^{-2}\) \(n =4\)
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