Question

For the cell reaction
\(2Fe^{3+}+2I^{-}_{(aq)}\rightarrow2Fe^{2+}+I_2{(aq)}\)
\(E^{\ominus}_{cell}=0.24\ V \ at \ 298 k.\)The standard Gibbs energy \((\Delta_rG^{\ominus})\) of the cell reaction is:
[Given that Faraday constant \( F=96500 \ C\ mol^{-1}]\)  

• A. \(-46.32 \ kjmol{-1}\)
• B. \(-23.16 \ kjmol{-1}\)
• C. \(46.32 \ kjmol{-1}\)
• D. \(23.16\ kjmol{-1}\)

Answer 1

The Correct Option is A

To find the standard Gibbs energy change \((\Delta_rG^{\ominus})\) for the given cell reaction, we can use the formula:

\[\Delta_rG^{\ominus} = -nFE^{\ominus}_{cell}\]

Where:

• \(n\) = number of moles of electrons transferred in the reaction
• \(F\) = Faraday constant = 96500 C mol^-1
• \(E^{\ominus}_{cell}\) = standard cell potential = 0.24 V

Step-by-step Calculation:

1. Identify the number of electrons transferred in the balanced reaction:

   The given reaction is:

   \(2Fe^{3+} + 2I^{-} \rightarrow 2Fe^{2+} + I_2\)

   Here, each Fe³⁺ ion gains 1 electron to become Fe²⁺. Therefore, for 2 Fe ions, 2 electrons are transferred.

   Thus, \(n = 2\).

2. Substitute the values into the equation for Gibbs energy:

   \[\Delta_rG^{\ominus} = -nFE^{\ominus}_{cell} = -(2 \times 96500 \times 0.24)\]

3. Calculate:

   \[\Delta_rG^{\ominus} = -2 \times 96500 \times 0.24 = -46320 \, \text{J mol}^{-1}\]

   Convert Joules to kilojoules:

   \[-46320 \, \text{J mol}^{-1} = -46.32 \, \text{kJ mol}^{-1}\]

Conclusion:

The standard Gibbs energy of the cell reaction is \(-46.32 \, \text{kJ mol}^{-1}\), hence the correct answer is the first option: \(-46.32 \, \text{kJ mol}^{-1}\).