Question

Consider the following reaction:
$Zn_{(s)} + Ag_2O_{(s)} + H_2O_{(l)} \rightarrow Zn^{2+}_{(aq)} + 2Ag_{(s)} + 2OH^-_{(aq)}$
Given: $E^\circ_{Ag^+/Ag} = 0.80~V$, $E^\circ_{Zn^{2+}/Zn} = -0.76~V$, $1~F = 96500~C~mol^{-1}$. $\Delta_r G^\circ$ for the above reaction is:

• A. $-301.080~kJ~mol^{-1}$
• B. $+310.080~kJ~mol^{-1}$
• C. $-326.070~kJ~mol^{-1}$
• D. $-375.060~kJ~mol^{-1}$

Hint:

Always ensure you convert the final answer from Joules to kiloJoules if the options demand it ($1~kJ = 1000~J$).

Answer 1

The Correct Option is A

Step 1: Concept
The standard Gibbs free energy change ($\Delta_r G^\circ$) of an electrochemical cell is related to its standard cell potential ($E^\circ_{cell}$) by the equation $\Delta_r G^\circ = -nFE^\circ_{cell}$.

Step 2: Meaning
Here, $n$ is the number of moles of electrons transferred in the balanced redox reaction, and $F$ is Faraday's constant.

Step 3: Analysis
1. Identify oxidation and reduction: Zinc goes from $0$ to $+2$ (oxidation, anode). Silver goes from $+1$ to $0$ (reduction, cathode).
2. Determine $n$: $Zn \rightarrow Zn^{2+} + 2e^-$, so $n = 2$.
3. Calculate $E^\circ_{cell}$: $E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80~V - (-0.76~V) = 1.56~V$.
4. Calculate $\Delta_r G^\circ$: $\Delta_r G^\circ = -2 \times 96500~C~mol^{-1} \times 1.56~V = -301080~J~mol^{-1}$.
5. Convert to kJ: $-301080~J~mol^{-1} = -301.080~kJ~mol^{-1}$.

Step 4: Conclusion
The calculated value perfectly matches option (A).

Final Answer: (A)