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Find the solubility of AgCl in 0.1 M CaC...

Find the solubility of `AgCl` in `0.1 M CaCl_(2)`. `E^(c-)._(Ag^(o+)|Ag)=0.799V` and that of `AgCl|Ag=0.222V`.

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To find the solubility of AgCl in a 0.1 M CaCl₂ solution, we will follow these steps: ### Step 1: Understand the Reaction AgCl dissociates in water as follows: \[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \] ### Step 2: Write the Expression for Ksp The solubility product (Ksp) expression for AgCl is given by: ...
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What would be the electrode potential of a silver electrode dipped in a saturated solution of AgCl in contact with 0.1 M KCl solution at 25^(@)C ? E^(c-)._(Ag^(o+)|Ag)=0.799 V K_(sp) of AgCl=1xx10^(-10)

The solubility of AgCl in 0.1M NaCI is (K_(sp) " of AgCl" = 1.2 xx 10^(-10))

Solubility of AgCl in 0.1 M CaCl_2 solution will be (K_(sp) of AgCl= 1.8×(10)^(-10) )

Calculate the potential of silver electrode in a saturated solution of AgBr(K_(sp)=6xx10^(-13)) containing 0.1 M KB r . E^(c-)._(Ag^(o+)|Ag)=0.80V.

Knowing that K_(sp) for AgCl is 1.0 xx 10^(-10) , calculate E for a silver // silver chloride electrode immersed in 1.00 M KCl at 25^(@)C.E^(c-)._(Ag^(o+)|Ag)=0.799V .

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+)^2]) . 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 26.5 13.24 53 106

The standard potential of the following cell is 0.23V at 15^(@)C and 0.21V at 35^(@)C: Pt|H_(2(g)|HCl(aq)|AgCl(s)|Ag(s) a. Write the cell reaction. b. Calculate DeltaH^(c-) and DeltaS^(c-) for the cell reaction by assuming that these quantities remain unchanged in the range 15^(@)C to 35^(@)C c. Calculate the solubility of AgCl in water at 25^(@)C . Given : The standard reduction potential of Ag^(o+)(aq)|Ag(s) is 0.80V at 25^(@)C .

The standard electrode of a metal ion (Ag|Ag^(o+)) and metal - insoluble salt anion (Ag|AgCl|Cl^(c-)) are related as (a) E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln K_(sp) (b) E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)\Ag)+(RT)/(F)lnK_(sp) (c) E^(c-)._(Ag^(o+)|Ag)=E^(c-)._(Cl^(c-)|AgCl|Ag)+(RT)/(F)ln.([Cl^(c-)])/(K_(sp)) (d) E^(c-)._(Cl^(c-)|AgCl|Ag)=E^(c-)._(Ag^(o+)|Ag)+(RT)/(F)ln.(K_(sp))/([Cl^(c-)])

The standard potential of the following cell is 0.23 V at 15^(@)C and 0.21V at 35^(@)C . Pt|H_(2) |HCl(aq) |Agcl (s)|g(s) (i) write the cell reaction . (ii) Calculate DeltaH^(@) and DeltaS^(@) for the cell reaction by assuming that these quantities remain unchanged in the range 15^(@)C to 35^(@)C . (iii) calculate the solubility of AgCl in water at 25^(@)C . Give , the standard reduction potential of the (Ag^(+)(aq) //Ag(s) is 0.80 V at 25^(@) C .

If solubility of AgCl(s) in water is 10^-5 M then the solubility of AgCl in 0.1 M AgNO_3 will be

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