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If `6.65xx10^(-2)g` of metalllic zinc is added to `100ml` saturated solution of `AgCl`, it react with `Ag^(+)` of solution as following reaction. `Zn(s)+2Ag^(+)(aq)hArrZn^(2+)(aq)+2Ag(s)` and approximately `10^(-x)` moles of `Ag` will be precipated. Calculate the value of `x` (Given `E_(Zn^(++)//Zn)^(@)=-0.76VE_(Ag^(+)//Ag)^(@)=0.8V, K_(sp)` of `AgCl=10^(-10)`, atomic mass of `Zn=65.3u, 10^(52.8813)=7.61xx10^(52)`) |
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Answer» Correct Answer - 6 `Zn(s)+2Ag^(+)(aq)hArrZn^(2+)(aq)+2Ag(s)` `E_("cell")^(@)=0.059/2log_(10)K_(eq)` `0.8+0.76=0.059/2 log_(10)K_(eq)` `K_(eq)=7.61xx10^(52)` As `K_(eq)` is very large the reactioin will tend towards completion Moles of `Ag^(+)` in solution `=(10^(-5)xx100)/1000=10^(-6)` As `Ag^(+)` is limiting reagent, the moles of `Ag` precipitate `~~10^(-6)` Hence `x=6` |
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