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a beaker contains 3 compartments separated by semipermeanble membrane left comp. contains Co(NO3)2 ,Eo(Co2+,Co)= --0.28,middle container contains KNO3 and the right comp. contains AgNO3 Eo=0.8, the conc. and vol. in each comp. is 0.1M,5 lit. and assume current is flowing in this electrolytic cell. find the con. of Co2+ and NO3-in the left compartment. plz show method also

Left compartment -

Co  Co²⁺ + 2e^-

Right compartment -

Ag  Ag⁺ + e^-

On adding these half cell reactions we get the cell reaction.

Co + 2Ag²⁺  2Ag + Co²⁺

E_cell = E^{0 _cell}  -  0.0591/n   * logCo/(Ag⁺ )²

E^{0 _cell} = E^{0 _{Ag+, Ag}}  -  E^{0 _{Co2+, Co}}

   = 0.8 - (-0.28)

   = 1.08

E_cell = 1.08 - 0.0591/2  log 0.1/(0.1)

   = 1.08 - 0.0591/2  log 10

   = 1.08 - 0.0295

   = 1.05