Question

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)` is
A. `2.14 v`
B. `4.28 v`
C. `6.42 v`
D. `8.56 v`

Answer 1

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`