Question

The standard free energy change ($\Delta G^\circ$) for the following reaction (in kJ) at 25°C is $$3Ca(s) + 2 Au^{+}(aq, 1M) \rightleftharpoons 3Ca^{2+}(aq, 1M) + 2Au(s)$$ (given: $E^\circ_{{Au}^{3+/2+}} = +1.50 \, V$, $E^\circ_{{Ca}^{2+/Ca}} = -2.87 \, V$, $F = 96500 \, {C mol}^{-1}$)

• A. $-2.53 \times 10^3$
• B. $+2.53 \times 10^3$
• C. $-2.53 \times 10^4$
• D. $+2.53 \times 10^4$

Hint:

The standard free energy change is related to the cell potential via the equation $\Delta G^\circ = -nFE^\circ$, where $n$ is the number of moles of electrons transferred in the reaction.

Answer 1

The Correct Option is A

The standard free energy change ($\Delta G^\circ$) for a reaction can be calculated using the following relation: $$\Delta G^\circ = -n F E^\circ$$ where: - $n$ is the number of moles of electrons transferred in the reaction,
- $F$ is the Faraday constant ($96500 \, {C mol}^{-1}$),
- $E^\circ$ is the cell potential.
Step 1: 
First, calculate the cell potential $E^\circ$. The overall cell reaction is: $${3Ca(s) + 2 Au}^{+}(aq, 1M) \rightleftharpoons 3Ca^{2+}(aq, 1M) + 2Au(s)$$ The standard cell potential $E^\circ_{{cell}}$ is given by: $$E^\circ_{{cell}} = E^\circ_{{cathode}} - E^\circ_{{anode}}$$ Here, the cathode is the reduction half-reaction involving ${Au}^{+}$, and the anode is the oxidation half-reaction involving ${Ca}$. Thus: $$E^\circ_{{cell}} = E^\circ_{{Au}^{3+/2+}} - E^\circ_{{Ca}^{2+/Ca}} = 1.50 \, {V} - (-2.87 \, {V}) = 4.37 \, {V}.$$ 
Step 2: Next, calculate the number of moles of electrons transferred in the reaction. From the balanced equation, we see that 6 electrons are transferred (3 moles of Ca give 6 electrons). 
Step 3: Now, we can calculate $\Delta G^\circ$: $$\Delta G^\circ = -6 \times 96500 \, {C mol}^{-1} \times 4.37 \, {V} = -2.53 \times 10^3 \, {kJ}.$$ Thus, the standard free energy change for the reaction is $-2.53 \times 10^3 \, {kJ}$.