On the basis of information available from the reaction

\(\frac{4}{3}Al + O_2 \rightarrow \frac{2}{3}Al_2O_3, \Delta G = -827 kJ /mol\) of O₂, the minimum emf required to carry out electrolysis of Al₂O₃ is

2.14 V

The correct answer is option 1. 2.14 V.

To calculate the minimum emf required for the electrolysis of \(Al_2O_3\), we can use the relationship between Gibbs free energy (\(\Delta G\)) and cell emf (\(E\)):

\(\Delta G = -nFE\)

where:
\(\Delta G\) is the Gibbs free energy change in Joules,
\(n\) is the number of moles of electrons transferred in the reaction,
\(F\) is the Faraday constant (96485 C/mol),
\(E\) is the cell emf in volts.

In the given reaction: \(\frac{4}{3}Al + O_2 \rightarrow \frac{2}{3}Al_2O_3\), 4 moles of electrons are transferred.

Given: \(\Delta G = -827 \, \text{kJ/mol} = -827000 \, \text{J/mol}\)

We need to convert the value of \(\Delta G\) to Joules and then calculate the cell emf (\(E\)).

\(\Delta G = -nFE\)
\(-827000 \, \text{J/mol} = -(4 \times 96485 \, \text{C/mol}) \times E\)
\(E = \frac{-827000 \, \text{J/mol}}{-(4 \times 96485 \, \text{C/mol})}\)
\(E = 2.14 \, \text{V}\)

Therefore, the correct answer is (1) 2.14 V.