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On the basis of information available from the reaction `(4)/(3)Al+O_(2) rarr (2)/(3)Al_(2)O_(3),DeltaG = -827kJ mol^(-1)` of `O_(2)`, the minimum emf required to carry out of the electrolysis of `Al_(2)O_(3)` is `(F=96,500 C mol^(-1))`A. `2.14 V`B. `4.28 V`C. `6.42 V`D. `8.56 V` |
Answer» Correct Answer - A We are given `(4)/(3) Al+O_(2) rarr (2)/(3) Al_(2)O_(3)` For every `1` mol of `O_(2)`, we need to perform electrolysis of `(2)/(3)` mol of `Al_(2)O_(3)`: `(2)/(3) mol Al_(2)O_(3) rarr (4)/(3) Al^(3+)+2O^(2-)` At cathode `(4)/(3)Al^(3+)+4e^(-) rarr (4)/(3) Al` At anode `2O^(2-) rarr 4e^(-)+O_(2)` Since `Delta_(r)G^(@) = -nFE_("cell")^(@)` we have `E_("cell")^(@) = (Delta_(r)G^(ɵ))/(-nF)` `= (-827 xx 10^(3) J mol^(-1))/(-(4 mol) (96,500 C mol^(-1)))` `= 2.14 V` |
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