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6.539 xx 10^(-2) of metallic Zn (65.39...

`6.539 xx 10^(-2)` of metallic Zn (65.39 amu) was added to 100 mL of saturated solution of AgCl. Calculate log `([Zn^(2+)] //[Ag^+]^2)`
Given `Ag^(+) + e = Ag , E^@ = 0.80 V`
`Zn^(2+) + 2e = Zn, E^@ = 0.76V`
Also find how many moles of Ag will be formed .

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`52.79 , 10^(-6)` mol
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6.537 xx 10^(-2)f of metallic Zn was added to 100 mL of saturated solution of AgCl . Calculate log .([Zn^(2+)])/([Ag^(o+)]) . Given : E^(c-)._(Ag^(o+)|Ag)=0.80V, E^(c-)._(Zn^(2+)|Zn)=-0.763V . K_(sp) of AgCl~~10^(-10), atomic weight of Zn=65.37

a. For the reaction Given : Ag^(o+)(aq)+Cl^(o+)(aq)rarr AgCl(s) Write the cell representation of above reaction and calculat E_("cell")^(@) at 298K. Also find the solubility product of AgCl. b. If 6.539xx10^(-2)g of metallic zinc is added to 100ml saturated solution of AgCl. Find the value of log_(10).([Zn^(2+)])/([Ag^(+)]^(2)) How many moles of Ag will be preciptated in the above reaction. Given that Ag^(+)+e^(-)rarrAg, E^(@)=0.80V Zn^(2+)+2e^(-)rarr Zn,E^(@)=-0.76V

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) )

For Zn^(2+) //Zn, E^@ =- 0. 76 V , for Ag^+ //Af E^@ =0.799 V . The correct statement is .

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