Questions & Answers

Question

Answers

A. \[4.5{\text{ }}V\]

B. \[3.0{\text{ }}V\]

C. \[ - 5.0\,\,V\]

D. \[ - 2.5\,\,V\]

Answer
Verified

Formula used: $\Delta G^\circ = - nFE{^\circ _{cell}}$

Where $\Delta G^\circ = $Gibbs free energy change

\[n = \] Number of electrons change

\[F = \] Faraday

\[E{^\circ _{cell}} = \] Potential difference

Relationship between Gibbs free energy change and the cell potential is given below

$\Delta G^\circ = - nF\,E{^\circ _{cell}}$………………….(i)

Where \[DG^\circ = \] Gibbs free energy change

\[n = \] number of electron transferred

\[F = \] Faraday

\[E{^\circ _{cell}} = \] potential difference

We know that electrolytic reduction of aluminum oxide

$\dfrac{2}{3}A{l_2}{O_3}\xrightarrow{{}}\dfrac{4}{2}Al + {O_2}\,\,\,\Delta G^\circ = + 960KJ/mol$

Number of electrons gain by metal $ = 4$ electrons

\[F = 96500\,C.\]

Substitute the value in Equation (1), we get

$960 \times 1000 = - 4 \times 96500 \times E{^\circ _{cell}}$, solve the equation and find the potential difference

$E{^\circ _{cell}} = - \dfrac{{960000}}{{4 \times 96500}}$

$ = - 2.48V$

So, potential difference is \[ - 2.5V\]

Hence, the correct option is D.