The standard free energy change (∆ {\G}^{\circ}. ) for the following reaction (in {\kJ} ) at 25^{\circ} {\C} is 3 {\Ca}( { s})+2 {\Au}^{3+}( {\aq}, 1 {\M}) \rightarrow 3 {\Ca}^{2+}( {\aq}, 1 {\M})+2 {\Au}( { s}) \text{ (given: } {\E}_{ {\Au}^{3+} \/ {\Au}}^0=+1.50 {\V}, {\E}_{ {\Ca}^{2+} \/ {\Ca}}^0=-2.87 {\V}, 1 {\F}=96500 {\C} {\mol}^{-1} \text{ ) } [AP EAPCET 21 May 2024 Shift 1]

Correct Answer is : −2.53×103

Thermodynamics

Section: Chemistry

Question : 24 of 34

Marks: +1, -0

The standard free energy change (∆G∘. ) for the following reaction (in kJ ) at 25∘C is
‌3‌Ca( s)+2Au3+(aq,1M)⟶3Ca2+(aq,1M)+2‌Au( s)

‌‌ (given: ‌EAu3+∕Au0=+1.50V,ECa2+∕Ca0=−2.87V,1F=96500Cmol−1‌ ) ‌

[AP EAPCET 21 May 2024 Shift 1]

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