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Step-by-Step Solution
Step 1: Identify the relevant equation
The relationship between standard Gibbs free energy change ( \Delta G^\circ ) and standard cell potential ( E_{cell}^\circ ) is given by:
\Delta G^\circ = -n F E_{cell}^\circ
Here,
• n = number of moles of electrons transferred
• F = Faraday constant
• \Delta G^\circ = standard Gibbs free energy change
• E_{cell}^\circ = standard cell potential
Step 2: Convert the given \Delta G^\circ into the appropriate units
The value of \Delta G^\circ is given as 17.37 kJ mol–1. We need it in joules per mole:
17.37\, \text{kJ mol}^{-1} = 17.37 \times 1000\, \text{J mol}^{-1} = 17370\, \text{J mol}^{-1}
Step 3: Plug in the known values into the equation
\Delta G^\circ = 17370\, \text{J mol}^{-1}, \quad
n = 3, \quad
F = 96500\, \text{C mol}^{-1}
\Delta G^\circ = -nFE_{cell}^\circ
Hence,
E_{cell}^\circ = -\frac{\Delta G^\circ}{n\,F}
Step 4: Calculate E_{cell}^\circ
E_{cell}^\circ = -\frac{17370}{3 \times 96500} \, \text{V}
Performing the calculation:
E_{cell}^\circ \approx -\frac{17370}{289500} \, \text{V} \approx -0.06 \, \text{V}
In scientific notation:
E_{cell}^\circ = -6.0 \times 10^{-2} \, \text{V}
Final Answer
The value of E_{cell}^\circ is
-6.0 \times 10^{-2} V.
