InterviewSolution
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InterviewSolution
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For an electrode reaction written as `M^(n+)+n e^(-)rarr M` , `E_("red")=E_("red")^(@)-(RT)/(nf)ln(1)/([M^(n+)])` `=E_("red")^(@)-(0.059)/(n)"llog"(1)/([M^(n+)])` at 298 k For the cell reaction `aA+bB rarr xX+yY` `rArr E_("cell")=E_("cell")^(@)-(RT)/(nf)"log"([X]^(x)[Y]^(y))/([A]^(a)[B]^(b))` For pure solids, liquids or gases at lampt. molarconc = 1 Std. free energy change `Delta G^(@)=-nfE^(@)` where f - for faraday = 96,500 c, n = no. of `e^(-)` `Delta G=-2.303 RT log K_(c )` where `K_(c )` is equilibrium constant `K_(c )` can be calculated by using `E^(@)` of cell. On the basis if information avaliable from the reaction `(4)/(3)Al+O_(2)rarr(2)/(3)Al_(2)O_(3) Delta G=-827 kJ//mol^(-1)` The minimum emf required to carry out an electrolysis of `Al_(2)O_(3)` isA. `2.14 v`B. `4.28 v`C. `6.42 v`D. `8.56 v` |
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Answer» Correct Answer - A For 1 mole of `O_(2)` `O_(2)rarr20^(-) i.e., 2//3xx30^(-)` or 4/3 mole of Al to change into `Al^(3+)` this n = 4 Thus `Delta G=-nFE` `E=-(Delta G)/(nF)=(-827000)/(4xx96500)=2.14v` |
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