Aconcentration cell is built based on the following half reactions by using two pieces of zinc as electrodes, two zn2+ solutions, 0.129 m and 0.427 m, and all other materials needed for a galvanic cell. what will the potential of this cell be when the cathode concentration of zn2+ has changed by 0.047 m at 291 k?
zn2+ + 2 e- ? zn eo = -0.761 v

Explanation:

The given reaction at cathode will be as follows.

At cathode: ,     = -0.761 V

At anode: ,       = 0.761

Therefore, net reaction equation will be as follows.

Initial:     0.129         -            -       0.427

Change:  -0.047      -            -     -0.047

Equilibrium: (0.129 - 0.047)      (0.427 - 0.047)

                = 0.082                       = 0.38

As for the given reaction is zero.

Hence, equation for calculating new cell potential will be as follows.

                   E_{cell} =

                              =

                              = 0.019

Thus, we can conclude that the cell potential of the given cell is 0.019.

a  solution  is always transparent, light passes through with no scattering from solute particles which are molecule in size. the  solution  is homogeneous and does not settle out. a colloid is intermediate between asolution  and a  suspension. while asuspension  will separate out a colloid will not.