= Solution
For the cell reaction normalized as in 4(b)(i), six moles of <electrons> are transferred per mole of reaction. Therefore
$$
\boxed{\Delta_rG=-6FE=-6(96485)(1.5180208)=-8.78797\times10^5\,\mathrm{J\,mol^{-1}}\approx-879\,\mathrm{kJ\,mol^{-1}}.}
$$
This is the <Gibbs energy of reaction> at the stated composition and gas pressure, not the <standard Gibbs energy of reaction>. Its negative sign confirms thermodynamic favourability in the written direction under those conditions.
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