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. The emf of the cell `Zn//zn^(2+)(0.01M)//Fe^(2+)(0.001M)//Fe` at 298 K is `0.2905`. The value of equilibrium constant for cell reaction siA. `e^((0.32)/(0.0295))`B. `1-^((0.32)/(0.0295))`C. `10^(0.26//0.0295)`D. `10^((0.32)/(0.059))` |
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Answer» Correct Answer - B `Zn+Fe^(2+)rarr Fe+Zn^(2+)` `E=E^(@)-(0.059)/(n)log Ke` Given E = 0.2905 i.e., `0.2905 = E^(@)-(0.059)/(2)"log"(0.01)/(0.001)` `E^(@)=0.2905+0.0295 log 10=0.32` `E^(@)=(0.059)/(n) log K_(eq) =0.32 =(0.059)/(2)log K_(eq)` `K_(eq)=10^(0.32//0.0295)` |
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