The magnitude of standard Gibbs free energy for the given cell reactio – InterviewSolution

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1.	The magnitude of standard Gibbs free energy for the given cell reaction in KJmol−1 at 298 K isZn(s)+Cu2+(aq)→Zn2+(aq)+Cu(s),E∘=2 V at 298 K (Faraday′s constant,F=96000 Cmol−1)
Answer» The magnitude of standard Gibbs free energy for the given cell reaction in $K J m o l^{- 1} a t 298 K$ is $Z n (s) + C u^{2 +} (a q) \to Z n^{2 +} (a q) + C u (s), E^{\circ} = 2 V a t 298 K$ $(F a r a d a y^{'} s c o n s t a n t, F = 96000 C m o l^{- 1})$

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