JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
Given the equilibrium constant \(K_c\) of the reaction \(Cu(s) + 2Ag{^+}_{(aq)} \to Cu^{+2}_{(aq)} + 2Ag(s)\) is \(10 \times 10^{15}\), calculate the \(E_{cell}^o\) of the reaction of \(298\,K\) [\({2.303\,\frac{{RT}}{F}}\) at \(298\,K\) \(=0.059\,V\)]
- A \(0.04736\,mV\)
- B \(0.4736\,mV\)
- C \(0.4736\,V\)
- D \(0.04736\,V\)
Answer & Solution
Correct Answer
(C) \(0.4736\,V\)
Step-by-step Solution
Detailed explanation
\(\Delta {G^o}\, = \, - \,nFE_{cell}^o\,\) \( = \, - \,2.303\,RT\,\log \,{K_{eq}}\) \(2FE_{cell}^o\, = \,2.303\,RT\,\log \,{K_{eq}}\) \(2E_{cell}^o\, = \,\frac{{2.303\,RT}}{F}\,\log \,10\, \times \,{10^{15}}\) \(E_{cell}^o\, = \,\frac{{0.059}}{2}\,\, \times \,16\)…
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