If a galvanic cell is produced using a nickel electrode and 1.0 M #Ni^(2+)# along with a silver electrode and 1.0 M #Ag^+#, calculate the concentration of #Ni^(2+)# and #Ag^+# in each compartment once the battery is dead?

sub-question: what does "once the battery is dead" mean?

1 Answer
Aug 2, 2018

Answer:

#sf([Ag^(+)])= # zero

#sf([Ni^(2+)]=1.5color(white)(x)"mol/l")#

Explanation:

A battery, or in this case, a cell, is "dead" when the potential difference between the two electrodes has fallen to zero.

Think of it like this: You turn on the tap in your kitchen and water flows out.

There are some who think that this is because there is a little guy from the water company out the back who works a little foot pump.

However, this is not the case. It flows out because there is a water tank somewhere in the building at a greater height - usually in the loft. Water in the tank has a greater potential energy so will naturally flow down to the kitchen.

If the tank was at the same height there would be no difference in potential energy so no water would flow.

It is like this in a cell. Current flows between the two electrodes if there is a difference in electrical potential between them.

This "potential difference" is measured in volts and is known as "The Electromotive Force" (emf). It is given the symbol #sf(E)#.

As you draw current from the cell during use, the potential difference between the two electrodes gradually falls and eventually becomes zero. The cell is "dead". I am sure you have noticed this with your cell phone.

In this case you can pump electrons back in and recharge the cell. These are called secondary cells. If not you have to re - cycle them and replace them with new ones.

You can't get something for nothing. (1st Law of Thermodynamics).

In this example the value of #sf(E_(cell)# has fallen to zero. The cell reaction has reached equilibrium:

#sf(Ni+2Ag^(+)rightleftharpoonsNi^(2+)+2Ag)#

Using published values:

#sf(E_(cell)^@=+0.8-(-0.25)=+1.05color(white)(x)V)#

The Nernst Equation at #sf(25^@C)# gives us :

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

#sf(z=2)#

When the cell is "dead" #sf(E_(cell)=0)# and so #sf(Q=K)#

#:.##sf(0=E_(cell)^@-(0.0591)/(2)logK)#

#sf(E_(cell)^@=(0.0591)/(2)logK=1.05)#

#sf(logK=35.058)#

#sf(K=1.14xx10^(35))#

This is such a fantastically high number that we can safely say that the reaction has gone to completion.

This means that, when the cell is "dead", the value of #sf([Ag^(+)]# has fallen to zero.

2 moles of #sf[Ag^(+)]# produce 1 mole of #sf(Ni^(2+)# so the final concentration of #sf(Ni^(2+))# will be #sf(1.0+0.5=1.5color(white)(x)"mol/l")#