© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Cell Reaction and Electron Transfer
The given overall cell reaction is:
$ \text{Cu} + 2\,\text{Ag}^+ \to \text{Cu}^{2+} + 2\,\text{Ag} $
From this balanced reaction, we see that 2 moles of electrons are transferred (since each Ag+ requires 1 electron for reduction, and there are 2 Ag+).
Step 2: Recall the Relationship Between $\Delta G^\circ$ and $E^\circ_{\text{cell}}$
The standard Gibbs free energy change $ \Delta G^\circ $ for an electrochemical cell is related to the standard cell potential $ E^\circ_{\text{cell}} $ by the formula:
$ \Delta G^\circ = -\,n\,F\,E^\circ_{\text{cell}} $
where
$ n $ is the number of moles of electrons exchanged in the reaction (here, $ n = 2 $).
$ F $ is the Faraday constant ($ F \approx 96500\,\text{C mol}^{-1} $).
$ E^\circ_{\text{cell}} $ is the standard cell potential (given as $ +\,0.46\,\text{V} $).
Step 3: Substitute the Known Values
Substitute $ n = 2 $, $ F = 96500\,\text{C mol}^{-1} $, and $ E^\circ_{\text{cell}} = 0.46\,\text{V} $ into the formula:
$ \Delta G^\circ = -\,2 \times 96500 \times 0.46 $
Calculate the product:
$ \Delta G^\circ = -\,88780 \,\text{J mol}^{-1} $
Since 1 kJ = 1000 J, we convert to kilojoules:
$ \Delta G^\circ = -\,88.78 \,\text{kJ mol}^{-1} \approx -\,89\,\text{kJ mol}^{-1} $
Step 4: State the Final Answer
Therefore, the standard Gibbs free energy change for the given reaction is:
$ \Delta G^\circ = -\,89\,\text{kJ mol}^{-1}. $
