Question

How amount of voltage should be supplieid to holfman voltameter in order to find the composition of water by volume and by mass?

Answer 1

You need to provide at least 2.07 V

Explanation:

Consider the standard electrode potentials:

`sf(2H^(+)+2erightleftharpoonsH_2color(white)(xxxxxxxxx)E^@=" "0.00color(white)(x)V)`

`sf(O_2+2H_2O+4erightleftharpoons4OH^(-)color(white)(xx)E^@=+1.23color(white)(x)V`

In a cell, the oxygen 1/2 cell has the more +ve value so this will take in electrons and shift to the right. The hydrogen electrode will take in these electrons and shift to the left.

The overall cell reaction will be:

`sf(O_2+2H_2O+2H_2rightleftharpoons4OH^(-)+4H^+)`

This becomes:

`sf(O_2+cancel(2)H_2O+2H_2rightleftharpoons+cancel(4)H_2O)`

`stackrel(color(white)(xxxxxxxxxx))(color(blue)(larr)" ""non-spontaneous"`

`sf(O_2+2H_2rightleftharpoons2H_2O)`

`stackrel(color(white)(xxxxxxxxxxxxxxxxxx))(color(red)(rarr)" ""spontaneous"`

`sf(E_(cell)^(@)=+1.23-0.00=+1.23color(white)(x)V)`

This is the basis of the hydrogen / oxygen fuel cell. The arrows indicate the direction of sponaneity.

Consider The Hofmann Voltameter:

In this process, electrical energy is put in to drive to reaction in the non - spontaneous direction:

`sf(2H_2Orarr2H_2+O_2)`

On this basis you would expect a voltage of at least 1.23 V would need to be applied to drive the reaction.

In practice, a little more will be needed. If you are familiar with chemical kinetics you will be aware that activation energy is needed to drive many reactions and that is one of the factors here.

The extra potential above the equilibrium potential of +1.23 V is called the polarization of the cell and is required to overcome the activation energy at the electrodes and in overcoming the resistance R in the electrolyte and leads.

The corresponding energy `sf(I^2R)` is dissipated as heat.

For platinum electrodes, which are usually used in The Hofmann Voltameter individual "overpotentials" for different electrodes have been measured:

`sf(H_2=+0.07color(white)(x)V)`

`sf(O_2=+0.77color(white)(x)V)`

To find the total overpotential of the cell you take the arithmetic difference between the two (`eta)`

In this case `sf(eta=+0.77-(-0.07)=+0.84color(white)(x)V)`

This is the extra voltage needed to drive the reaction.

The total voltage will be:

`sf(DeltaPhi=1.23+0.84=+2.07color(white)(x)V)`

For a spontaneous process, such as the fuel cell, the maximum work you can get from the cell is given by:

`sf(DeltaG^@=-nFE^@)`

You would need to subtract 0.84 V from the value of `sf(E^@)` to get the actual voltage such that:

`sf(DeltaPhi=1.23-0.84=+0.39color(white)(x)V)`