A concentration cell similar to the onse shown is composed of teo Zn electrodes and solutions of different Zn2+ concentrations. The left compartment contains .565 M Zn2+, and the right compartment contains 0.319M Zn2+. Calculate the cell potential for this reaction at 298K.

In this zinc concentration cell the reaction would proceed spontaneously _(from left to right, from right to left?)

The cell emf for this cell is 0.077 V

Explanation:

The balanced cell reaction will be,  

Oxidation half reaction (anode):  

Reduction half reaction (cathode):  

In this case, the cathode and anode both are same. So,  is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

where,

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

= ?

= 0.0111 M

= 4.50 M

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

Therefore, the cell emf for this cell is 0.077 V

i think its 100.

hope this !