If the concentration of sn2+ in the cathode compartment is 1.40 m and the cell generates an emf of 0.25 v , what is the concentration of pb2+ in the anode compartment?

Answer : The concentration of in the anode compartment will be,

Solution :

The balanced cell reaction will be,  

Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Tin (Sn) undergoes reduction by gain of electrons and thus act as cathode.

First we have to calculate the standard electrode potential of the cell.

Now we have to calculate the concentration of

Using Nernest equation :

where,

n = number of electrons in oxidation-reduction reaction = 2

= emf of the cell = 0.25 V

Now put all the given values in the above equation, we get  concentration of

Therefore, the concentration of in the anode compartment will be,

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

c. 4.54 x 10–12 j happens to be correct