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: Key formula.
The standard Gibbs free energy change is related to the standard cell potential by $\Delta_r G^\circ = -nFE^\circ_{cell}$, where $n$ is the number of moles of electrons transferred and $F = 96500~C~mol^{-1}$.
Step 2: Identify the half-reactions.
Zinc is oxidised at the anode: $Zn \rightarrow Zn^{2+} + 2e^-$. Silver ion is reduced at the cathode: $2Ag^+ + 2e^- \rightarrow 2Ag$. Therefore $n = 2$.
Step 3: Calculate $E^\circ_{cell}$.
$E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} = 0.80 - (-0.76) = 1.56~V$.
Step 4: Calculate $\Delta_r G^\circ$.
\[ \Delta_r G^\circ = -2 \times 96500 \times 1.56 = -301080~J~mol^{-1} = -301.080~kJ~mol^{-1} \]
\[ \boxed{(A)} \]