JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
If the standard electrode potential for a cell is \(2\,V\) at \(300\,K,\) the equilibrium constant \((K)\) for the reaction \(Zn\,(s)\,\, + \,\,C{u^{2 + }}\,(aq)\, \rightleftharpoons \,Z{n^{2 + \,}}\,(aq)\, + Cu\,(s)\) At \(300\,K\) is approximately \((R = 8 \,JK^{-1}\,mol^{-1},\, F = 96000\,C\,mol^{-1})\)
- A \(e^{-80}\)
- B \(e^{-160}\)
- C \(e^{320}\)
- D \(e^{160}\)
Answer & Solution
Correct Answer
(D) \(e^{160}\)
Step-by-step Solution
Detailed explanation
\(Zn\,(S)\, + \,C{u^{2 + }}(aq)\, \rightleftharpoons \,Z{n^{2 + }}\,(aq)\, + \,Cu(s)\) \( - nF{E_{cell}}\, = \, - \,RT\ln \,K\) \(\ln \,K\, = \,\frac{{2\, \times \,96500\, \times \,2}}{{8\, \times \,300}}\, = \,160.83\) \(K\, = \,{e^{160}}\)
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