JEE Mains · Chemistry · STD 12 - 2. Electrochemistry
The standard Gibbs energy for the given cell reaction in \(kJ\,mol^{-1}\) at \(298\,K\) is \(Zn(s)\, + \,C{u^{2 + }}(aq)\, \to \,Z{n^{2 + }}(aq) + Cu\,(s)\,,\,{E^o} = 2\,V\) at \(298\,K\) [Faraday’s constant \(F = 96500\,C\,mol^{-1}\) ]
- A \(-192\)
- B \(384\)
- C \(-384\)
- D \(192\)
Answer & Solution
Correct Answer
(C) \(-384\)
Step-by-step Solution
Detailed explanation
\(\Delta G = - 2 \times 96000 \times 2\) \(\Delta G = - nF{E^o}\) \(\Delta G = - 2 \times 96000 \times 2\,J\) \(\Delta G = - 384\,kJ\,mol^{-1}\)
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