Home
Class 12
CHEMISTRY
The solubility product of Pb(3)(AsO(4))(...

The solubility product of `Pb_(3)(AsO_(4))_(2)` is `4.1xx10^(-36). E^(c-)` for the reaction `:`
`Pb_(3)(AsO_(4))_(2)(s)+6e^(-)hArr3Pb(s)+2AsO_(4)^(2-)`
`E_((Pb)2^(+)|Pb)^(Θ)=-0.13V`

A

`+0.478V`

B

`-0.13V`

C

`-0.478V`

D

`+0.13V`

Text Solution

Verified by Experts

The correct Answer is:
c

`E^(c-)._(AsO_(4)^(2-)|Pb_(3)(AsO_(4))^(2)|Pb)=E^(c-)._(Pb^(2+)|Pb)+(0.059)/(6)logK_(sp)`
`=-0.13+(0.059)/(6)log4.1xx10^(-36)`
`=-0.13-0.348=-0.478V`
Promotional Banner

Similar Questions

Explore conceptually related problems

Complete the equation: Pbs+4O_3rarr……..+4O_2 .

The equilibrium constant of the reaction : A_((s))+2B_((aq))^(+)hArr A_((aq))^(2+)+2B_((s)): E_(cell)^(@)=0.0295V is :

Consider the following half reactions : PbO_(2)(s)+4H^(o+)(aq)+SO_(4)^(2-)+2e^(-)rarrPbSO_(4)(s)+2H_(2)O" "E^(c-)=+1.70V PbSO_(4)(s)+2e^(-)rarr Pb(s)+SO_(4)^(2-)(aq)" "E^(c-)=-0.31V a. Calculate the value of E^(c-) for the cell. b. Calculate the voltage generated by the cell if [H^(o+)]=0.10M and [SO_(4)^(2-)]=2.0M c. What voltage is generated by the cell when it is at chemical equilibrium ?

a) Complete the following equations: i) PbS_((s))+4O_(3(g))rarr ii) 2NaOH+SO_(2)rarr

2Pb(NO_(3))_(2) to 2PbO+nA+O_(2) . What is nA in the given reaction ?

The balanced chemical equation for the reaction : MnO_(4)^(-)+AsO_(3)^(3-)+H^(+)toMn^(2+)+AsO_(4)^(3-)+H_(2)O involves

Pb_(3)O_(4)+4HNO_(3)to2Pb(NO_(3))_(2)+PbO_(2)+2H_(2)O Which is true about the lead ions in Pb_(3)O_(4) ?