Question #867f3

1 Answer
May 16, 2017

Answer:

#sf(E_(cell)=+0.58color(white)(x)V)#

Explanation:

#sf(Ni^(2+)+2erightleftharpoonsNi" "E^@=-0.25color(white)(x)V)#

#sf(Cu^(2+)+2erightleftharpoonsCu" "E^@=+0.34color(white)(x)V)#

The nickel 1/2 cell has the more -ve electrode potential so will push out electrons and shift right to left.

The copper 1/2 cell will take in these electrons and shift left to right.

This gives the overall cell reaction:

#sf(Ni+Cu^(2+)rarrNi^(2+)+Cu)#

To find #sf(E_(cell)^@)# subtract the least +ve value from the most +ve:

#sf(E_(cell)^@=+0.34-(-0.25)=+0.59color(white)(x)V)#

Now we need to use the Nernst Equation:

#sf(E_(cell)=E_(cell)^@-(RT)/(zF)lnQ)#

At 298K this can be written:

#sf(E_(cell)=E_(cell)^@-(0.0591)/(z)logQ)#

#sf(z)# is the no. moles of electrons transferred which, in this case = 2.

Q is the reaction quotient and from the cell equation is given by:

#sf(Q=a_(Ni^(2+))/(a_(Cu^(2+)))#

Putting in the numbers:

#sf(E_(cell)=+0.59-0.0591/(2)log[0.05/0.025])#

#sf(E_(cell)=+0.59-0.008895color(white)(x)V)#

#sf(E_(cell)=+0.58color(white)(x)V)#