Question

The standard Gibbs free energy change \( \Delta G^\circ \) of a cell reaction is \(-301 { kJ/mol}\). What is \( E^\circ \) in volts? 
(Given: \( F = 96500 { C/mol}\), \( n = 2 \))

• A. \(2.56\)
• B. \(-1.56\)
• C. \(1.20\)
• D. \(1.56\)

Hint:

The sign of \( \Delta G^\circ \) affects the polarity of \( E^\circ \); a negative \( \Delta G^\circ \) leads to a positive \( E^\circ \), indicating a spontaneous reaction.

Answer 1

The Correct Option is D

The relationship between the standard Gibbs free energy change (\(\Delta G^\circ\)) and the standard cell potential (\(E^\circ\)) is given by the equation: \[ \Delta G^\circ = -nFE^\circ \] where:

• \(\Delta G^\circ\) is the standard Gibbs free energy change (in Joules)
• \(n\) is the number of moles of electrons transferred in the reaction
• \(F\) is Faraday's constant (\(96500 \, \text{C/mol}\))
• \(E^\circ\) is the standard cell potential (in Volts)

Given values:

• \(\Delta G^\circ = -301 \, \text{kJ/mol} = -301 \times 10^3 \, \text{J/mol}\)
• \(F = 96500 \, \text{C/mol}\)
• \(n = 2\)

Step 1: Solve for \(E^\circ\) Rearranging the equation: \[ E^\circ = -\frac{\Delta G^\circ}{nF} \] Step 2: Substituting the given values: \[ E^\circ = -\frac{-301 \times 10^3 \, \text{J/mol}}{2 \times 96500 \, \text{C/mol}} \] \[ E^\circ = \frac{301 \times 10^3}{2 \times 96500} \, \text{V} \] \[ E^\circ = \frac{301000}{193000} \, \text{V} \] \[ E^\circ = \frac{3010}{1930} \, \text{V} \] \[ E^\circ = \frac{301}{193} \, \text{V} \] \[ E^\circ \approx 1.559585... \, \text{V} \] Step 3: Rounding to two decimal places: \[ E^\circ \approx 1.56 \, \text{V} \] Correct Answer: (4) 1.56