CBSE 2025, Set 2

Question

(a) Calculate the standard Gibbs energy ($\Delta_r G°$) of the following reaction at 25°C:

$	ext{Au(s)} + 	ext{Ca}^{2+}(1	ext{M}) ightarrow 	ext{Au}^{3+}(1	ext{M}) + 	ext{Ca(s)}$

$E°_{	ext{Au}^{3+}/	ext{Au}} = +1.5 	ext{ V},\quad E°_{	ext{Ca}^{2+}/	ext{Ca}} = -2.87 	ext{ V}$

Predict whether the reaction will be spontaneous or not at 25°C.

[1 F = 96500 C mol$^{-1}$]

Edvaya · Accepted Answer

The correct answer is Subjective. **Step 1:** Identify anode and cathode. Au is oxidised (anode): Au → Au³⁺ + 3e⁻ Ca²⁺ is reduced (cathode): Ca²⁺ + 2e⁻ → Ca To balance electrons (LCM of 3 and 2 = 6): $$2	ext{Au} + 3	ext{Ca}^{2+} ightarrow 2	ext{Au}^{3+} + 3	ext{Ca} \quad (n = 6)$$ **Step 2:** Calculate $E°_{cell}$ $$E°_{cell} = E°_{cathode} - E°_{anode} = -2.87 - (+1.5) = -4.37 	ext{ V}$$ **Step 3:** Calculate $\Delta_r G°$ $$\Delta_r G° = -nFE°_{cell} = -(6)(96500)(-4.37)$$ $$= 6 	imes 96500 	imes 4.37 = 2{,}530{,}230 	ext{ J mol}^{-1}$$ $$= \boxed{2530.23 	ext{ kJ mol}^{-1}}$$ **Step 4:** Predict spontaneity $\Delta_r G° = +2530.23$ kJ mol⁻¹ > 0 and $E°_{cell} = -4.37$ V < 0 **The reaction is non-spontaneous** at 25°C under standard conditions.

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