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 determine the standard Gibbs energy change, \(\Delta_rG^{\ominus}\), of the cell reaction, we use the following formula:

\(\Delta_rG^{\ominus} = -nFE^{\ominus}_{cell}\)

Where:

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

In the given reaction:

\(2Fe^{3+}+2I^{-}_{(aq)}\rightarrow2Fe^{2+}+I_2{(aq)}\)

The number of electrons transferred is 2. Therefore, \(n = 2\).

Now, substitute the values into the formula:

\(\Delta_rG^{\ominus} = -2 \times 96500 \times 0.24 \ V\)

\(\Delta_rG^{\ominus} = -46240 \ J \ mol^{-1}\)

Convert joules to kilojoules:

\(\Delta_rG^{\ominus} = -46.24 \ kJ \ mol^{-1}\)

Therefore, the standard Gibbs energy of the cell reaction is approximately \(-46.32 \ kj/mol\).

Hence, the correct answer is \(-46.32 \ kj/mol\).