Question

Galvanic cell potential ??

At 25 ºC, a galvanic cell is prepared with a metallic lead electrode submerged in
250 mL of an aqueous solution containing 41.4 g of lead nitrate (II) and with a
metallic iron electrode submerged in 1 L of an iron (II) sulphate solution.
The concentration of the iron (II) sulphate solution can be determined by the
titration reaction that requires 20 mL of potassium permanganate solution 0.1
M to completely oxidize to Fe3 + (ac) the Fe2 + (ac) that is in 25 mL of the solution of
iron sulfate (II). In this reaction, the manganese is reduced to Mn2 + (ac). Calculate:
a) The initial potential of the stack.
b) The potential of the stack when equilibrium is reached.
c) The value of the equilibrium constant.
d) The [Fe2 +] and [Pb2 +] when the battery has produced a current of 10 amps
for half an hour.
e) The potential of the battery once the amount of electricity has been produced
corresponding to section d).
f) The variation of mass experienced by the iron electrode.

Answer 1

Long question = long answer.

See below:

Explanation:

(a).

We need to find the concentration of the iron(II) and lead(II) ions.

`sf(MnO_4^(-)+8H^(+)+5Fe^(2+)rarrMn^(2+)+5Fe^(3+)+4H_2O)`

`:.`1 mole `sf(MnO_4^(-)-=)`5 mole `sf(Fe^(2+))`

`sf(n_(MnO_4^(-))=0.1xx20/1000=0.002)`

`:.` `sf(n_(Fe^(2+))=0.002xx5=0.01)`

`sf(c=n/v)`

`:.` `sf([Fe^(2+)]=0.01/0.025=0.4color(white)(x)"mol/l")`

For lead(II) nitrate `sf(M_r=331.20)`

`:.` `sf(n_(Pb^(2+))=41.4/331.2=0.125)`

`:.` `sf([Pb^(2+)]=0.125/0.250=0.500color(white)(x)"mol/l")`

Now we use standard electrode potentials to find `sf(E_(cell)^@)`:

`sf(Fe^(2+)+2erightleftharpoonsFecolor(white)(xxxxxx)E^@=-0.44V)`

`sf(Pb^(2+)+2erightleftharpoonsPbcolor(white)(xxxxxx)E^@=-0.126V)`

This tells us that the spontaneous cell reaction is:

`sf(Fe+Pb^(2+)rarrFe^(2+)+Pb)`

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

`sf(E_(cell)^@=-0.126-(-0.44)=+0.314color(white)(x)V)`

At 298K the Nernst Equation can be written:

`sf(E_(cell)=E_(cell)^@-0.0591/(2)logcolor(white)(x)([Fe^(2+)])/([Pb^(2+)]))`

`sf(E_(cell)=0.314-0.0591/(2)log[0.4/0.5]color(white)(x)V)`

`sf(E_(cell)=0.314+0.0028637color(white)(x)V)`

`sf(color(red)(E_(cell)=0.317color(white)(x)V)`

(b).

When equilibrium has been reached the potential difference between the two 1/2 cells falls to `sf(color(red)(zero))`.

(c).

At equlibrium we have:

`sf(E_(cell)=E^@-0.0591/(2)logK)`

Since `sf(E_(cell)=0)` this becomes:

`sf(E^@=0.0591/(2)logK)`

`sf(logK=(2E^@)/(0.0591)=(2xx0.314)/(0.0591)=10.626)`

From which `sf(color(red)(K=4.23xx10^(10))`

(d).

`sf(color(white)(xxxxxxxxxxxx)Fe+Pb^(2+)rarrFe^(2+)+Pb)`

Initial mol.`sf(color(white)(xxxxxxxxx)0.125color(white)(xxx)0.01)`

As current is drawn from the cell the no. moles `sf(Pb^(2+))` goes down and the no. moles `sf(Fe^(2+))` goes up by the same amount.

10 A is a huge current to draw from such a cell, anyway:

`sf(Q=It=10xx30xx60=18,000color(white)(x)C)`

The charge on a mole of electrons is The Faraday.

1F = `sf(9.65xx10^(4)color(white)(x)"C/mol")`

`:.` No. of Faradays passed = `sf((1.8xx10^(4))/(9.65xx10^(4))=0.1865color(white)(x)F)`

Because we have 2+ ions then:

2F will discharge 1 mole `sf(Pb^(2+))`

`:.` `sf(0.1865Frarr0.1865/(2)=0.09326)` moles of `sf(Pb^(2+))`

This means the no. moles of `sf(Pb^(2+)` left = 0.125 - 0.09326 = 0.03174 mol.

`:.` `sf([Pb^(2+)]=0.03174/(0.250)=color(red)(0.12696color(white)(x)"mol/l")`

The no. of moles of `sf(Fe^(2+))` will have increased by this amount:

`sf(n_(Fe^(2+))=0.01+0.09326=0.10326)`

`:.` `sf([Fe^(2+)]=0.10326/(0.025)=color(red)(4.1304color(white)(x)"mol/l"`

(e).

`sf(E_(cell)=E_(cell)^@-0.0591/(2)log[4.1304/0.12696]color(white)(x)V)`

`sf(E_(cell)=0.314-0.0591/(2)xx1.5123color(white)(x)V)`

`sf(color(red)(E_(cell)=0.2693color(white)(x)V)`

As you can see the emf of the cell has fallen.

(f).

The iron anode loses 0.0936 mol.

`sf(A_r=55.845)`

`:.` `sf(m_(Fe)=0.09326xx55.845=color(red)(5.208color(white)(x)g)`

This is the loss of mass of the iron anode.