The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`

The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C`

1.	The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follow : `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2), Delta_(r)G= +960 kJ mol^(-1)` The potential difference needed for the electrolytic reduction of aluminium oxide `(Al_(2)O_(3))` at `500^(@)C` isA. `5.0 V`B. `4.5 V`C. `3.0 V`D. `2.5 V`
Answer» Correct Answer - D We are given decomposition reaction: `(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al+O_(2)` The electrolyte `Al_(2)O_(3)` splits into ions as follows: `Al_(2)O_(3) rarr 2Al^(3+)+3O^(2-)` At cathode `2Al^(3+)+6e^(-) rarr 2Al` At anode `3O^(2-) rarr 6e^(-)+(3)/(2)O_(2)` We have `Delta_(r)G^(ɵ) = -nfE_("cell")^(ɵ)` or `E_("cell")^(ɵ) = (Delta_(r)G^(ɵ))/(-nF) = (960 xx 10^(3)J mol^(-1))/(-(6 mol)(96500 C)) = -1.66 V`

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