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AP EAMCET · Chemistry · Electrochemistry

The standard free energy change \(\left(\Delta G^{\circ}\right)\) for the following reaction (in kJ ) at \(25^{\circ} \mathrm{C}\) is \(3 \mathrm{Ca}(\mathrm{s})+2 \mathrm{Au}^{3+}\) (aq. 1 M ) \(\rightarrow\) \(3 \mathrm{Ca}^{2+}(\mathrm{aq}, 1 \mathrm{M})+2 \mathrm{Au}(\mathrm{s})\)
(given : \(\mathrm{E}_{\mathrm{Au}^{3+} / \mathrm{Au}}^{\mathrm{o}}=+1.50 \mathrm{~V}, \mathrm{E}_{\mathrm{Ca}^{2+} / \mathrm{Ca}}^0=-2.87 \mathrm{~V}\), \(1 \mathrm{~F}=96500 \mathrm{C} \mathrm{mol}^{-1}\) )

  1. A \(-2.53 \times 10^3\)
  2. B \(+2.53 \times 10^3\)
  3. C \(-2.53 \times 10^4\)
  4. D \(+2.53 \times 10^4\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(-2.53 \times 10^3\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & \text { } 3 \mathrm{Ca}(\mathrm{~s})+2 \mathrm{Au}^{3+}(\mathrm{aq} 1 \mathrm{M}) \rightarrow 3 \mathrm{Ca}^{2+}(\mathrm{aq} \mathrm{1M})+2 \mathrm{Au}(\mathrm{~S}) \\ & \mathrm{E}^0=\mathrm{E}_{\mathrm{L}}+\mathrm{E}_{\mathrm{R}} \\ & =\mathrm{E}_{\mathrm{Ca}…

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