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,

the density(ratio between mass and volume as m/v) decreases from top to bottom in periodic table group as the size increases and nuclear charge decreases. but in periods density increases from left to right as atom size decreases and nuclear charge increases.