Consider the following reaction proceeding at 298.15 k: cu(s)+2ag+(aq,0.15 m)⟶cu2+(aq, 1.14 m)+2ag(s) if the standard reduction potential for copper(ii) is 0.34 v and the standard reduction potential for silver(i) is 0.80 v what is the cell potential for this cell, in volts

Answer : The cell potential for this cell 0.434 V

Solution :

The balanced cell reaction will be,  

Here copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) 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 cell potential for this cell.

Using Nernest equation :

where,

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

= ?

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

Therefore, the cell potential for this cell 0.434 V

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

explanation:

wheres the pic