Question

Reduction Half Reaction E (V)

Ag₂MoO₄(s) + 2e^- ---> 2 Ag(s) + MoO₄^2-(aq) 0.4573 V

Ag⁺(aq) + e^- ---> Ag(s) 0.7996 V

a.) Calculate the mass in grams of Ag₂MoO₄(s) that will dissolve in 2.0 L of water.

b.) Calcilate the cell potential of:

Ag(s) | Ag₂MoO₄(s) | MoO₄^2-(aq) (0.010 M) || Ag⁺(aq) (0.010M) | Ag (s)

Answer 1

a)

Ag₂MoO₄(s) + 2e^- ---> 2 Ag(s) + MoO₄^2-(aq) Eo= 0.4573 V [Reduction]

2Ag(s) ----> 2Ag⁺(aq) +2 e^- Eo=-0.7996V (reversed reaction)[oxidation]

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Add:Ag₂MoO₄(s)----> 2Ag⁺(aq) +MoO₄^2-(aq)   Eocell=Eored+Eo(ox)=0.4573-0.7996=-0.342V

Solubility reaction:Ag₂MoO₄(s)----> 2Ag⁺(aq) +MoO₄^2-(aq) Eo=-0.342V

Ksp=Keq=equilibrium constant=solubility product

Using equation for free energy, Go=-nFEo ,n=mol of electrons exchanged,Eo=std reduction potential,

F=faraday's constant=96485 C/mol

Also,Go=-RT lnKeq

where R=Universal gas constant=8.314J/K.mol

T=temperature=298K

Keq=equilibrium constant for the reaction

Go=-RT lnKeq=-nFEo

or, ln Ksp=nFEo/RT=2 mol electrons*(96485 C/mol)*(-0.342J/C)/(8.314J/Kmol)*298K=-26.66

ksp=exp(-26.66)=2.639*10^-12

Also Ksp=[Ag+]^2 [MoO42-]

Let the solubility of Ag+ be S,so [Ag+]=S,[MoO42-]=S/2

Ksp=(S)^2*(S/2)=0.5 S^3

2.639*10^-12=0.5 S^3

S=1.741*10^-4M=solubility of Ag₂MoO₄(s)

So ,1.741*10^-4 mol of Ag₂MoO₄(s) will dissolve in 1L

2*1.741*10^-4 mol of Ag₂MoO₄(s) will dissolve in 2L

or ,molar mass of Ag₂MoO₄(s)*2*1.741*10^-4 mol Ag₂MoO₄(s)=mass of Ag₂MoO₄(s) will dissolve in 2L water

or,375.67 g/mol*2*1.741*10^-4 mol=0.131g will dissolve in 2L water

Answer:0.131g

b)Using Nerst equation,

E=Eocell-RT lnQ

Q=[Ag⁺]/[MoO₄^2-(aq)]=0.01M/0.01M=1

So,E=Eocell=Eo red+Eo ox

As cathode is ,Ag⁺(aq) + e^- ---> Ag(s) Eo(red)=0.7996V

Eo(Ox)=0.4573V

Ecell=0.7996-0.4573=0.342