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 VB. 4.28 VC. 6.42 VD. 8.56 V |
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Answer» Correct Answer - A `Al rarr Al^(3+)+3e^(-)` `4/3 mol Al equiv 4/3 xx 3 mol e^(-)` `equiv 4 mol e^(-)` i.e., `n=4` `DeltaG=-nFE` `-827xx1000=-4xx96500xxE` |
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