`EMF` of the cell
`|Ag|AgNO_(3)(0.1M)||K Br(1 N),AgBr(s)|Ag is -0.6V` at `298K`
`AgNO_(3) ` is `80%` and `KBr` is `60%` dissociated.
Calculate `a.` Solubility and
`b. K_(sp)` of `AgBr` at `298 K`.

The EMF of the cell Ag|0.1(N)AgNO_(3)||1(N) KBr, AgBr(s)|Ag was found to be -0.64V at 298 K. 0.1 N Ag NO_(3) is 81.3% dissociated and 1N KBr is 75.5% dissociated. Calculate : a. Solubility b. Solubility product of AgBr at 298 K

The EMF of the cell : Ag|AgCl,0.1 MKCl||0.1 M AgNO_(3)|Ag is 0.45V. 0.1 M KCl is 85% dissociated and 0.1 M AgNO_(3) is 82% dissociated. Calculate the solubility product of AgCl at 25^(@)C .

The EMF of the cell : Ag|AgCl,0.1 MKCl||0.1 M AgNO_(3)|Ag is 0.45V. 0.1 M KCl is 85% dissociated and 0.1 M AgNO_(3) is 82% dissociated. Calculate the solubility product of AgCl at 25^(@)C .

The EMF of the cell : Ag|Ag_(2)CrO_(4)(s),K_(2)CrO_(4)(0.1 M)||AgNO_(3)(0.1M)||Ag is 206.5mV. Calculate the solubility of Ag_(2)CrO_(4) in 1M Na_(2)CrO_(4) soluion.

The EMF of the cell : Ag|Ag_(2)CrO_(4)(s),K_(2)CrO_(4)(0.1 M)||AgNO_(3)(0.1M)||Ag is 206.5mV. Calculate the solubility of Ag_(2)CrO_(4) in 1M Na_(2)CrO_(4) solution.

Calculate the emf of the following cell at 25^(@)C Ag(s)|AgNO_(3) (0.01 M)||AgNO_(3) (0.05 M)|Ag (s)