1.

Calculate the EMF of the cell for the reaction. Mg_((s))+2Ag_((aq))^+ Mg_((aq))^++2Ag(s) Given: E^@ Mg^(2+)//Mg=-2.37 V E^@ Ag^(+)//Ag=0.08V [Mg^(2+)]=0.001M,[Ag^+]=0.0001M log 10^5=5

Answer»

Solution :(a) At anode: `Pb(s) + SO_(4)^(2-) (aq) to PbSO_4(s) + 2e^(-)`
At cathode : `PbO_2(s) + SO_(4)^(2-)(aq) + 4H^(+) (aq0 + 2e^(-) to PbSO_4(s) + 2H_2O(1)`
(b) `DeltaG^0 = -nFE_("cell")^(0)`
`N = 6`
`F = 96487 c//mol`
`E_("cell")^(0) = E_(Al)^(0) - E_(MG)^(0) = -1.66V - (-2,36 V) = 0.7V`
`Delta G^(0) = -6 xx 96487 xx 0.7= -405245 J//mol OR -405.24 kJ//mol`.


Discussion

No Comment Found

Related InterviewSolutions