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The concentration of Ag2 Cr2O7 is 6.5xx1...

The concentration of `Ag_2 Cr_2O_7` is `6.5xx10^(-5)` M in concentrated solution of `Cr_2O_7^(2-)` at a temp. then calculate `K_2Cr_2O_7`

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`K_(sp)=[Ag^+]^2 [Cr_2O_7^(2-)]`
`therefore K_(sp)=4S^3=4(6.5xx10^(-5))^3`
`=1.098xx10^(-12)`
Where S=Solubility of `Ag_2Cr_2O_7=6.5xx10^(-5)` M
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A half cell is prepared by K_(2)Cr_(2)O_(7) in a buffer solution of pH =1 . Concentration of K_(2)Cr_(2)O_(7) is 1M . To 3 litre of this solution 570 gm of SnCI_(2) is added which is oxidised completely to SnCI_(4) . Given: E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06 , Atomic of mass Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15 Half cell potential E_(Cr_(2)O_(7)^(-2)//Cr^(+3))^(@) after teh reaction of SnCI_(2) is:

A half cell is prepared by K_(2)Cr_(2)O_(7) in a buffer solution of pH =1 . Concentration of K_(2)Cr_(2)O_(7) is 1M . To 3 litre of this solution 570 gm of SnCI_(2) is added which is oxidised completely to SnCI_(4) . Given: E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06 , Atomic of mass Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15 Emf of the cell Pt|underset((0.1M))(Sn^(+2)),underset((0.2M))(Sn^(+4)),underset((1M))(H^(+)):||:underset((0.2M))(Cr_(2)O_(7)^(2-)),underset((1M))(Cr^(+3)),underset((1M))(H^(+)):|:Pt

A half cell is prepared by K_(2)Cr_(2)O_(7) in a buffer solution of pH =1 . Concentration of K_(2)Cr_(2)O_(7) is 1M . To 3 litre of this solution 570 gm of SnCI_(2) is added which is oxidised completely to SnCI_(4) . Given: E_(Cr_(2)O_(7)^(-2)//Cr^(+3).H^(+))^(@) = 1.33V, (2.303)/(F) RT = 0.06 , Atomic of mass Sn = 119, E_(Sn^(+4)//Sn^(+2))^(@) = 0.15 Number of moles of Cr^(+3) formed are

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