Question

The Ksp of copper(II) ferrocyanide (Cu2[Fe(CN)6]) is 1.3 × 10−16 at 25°C. Determine the potential of a concentration cell in which one half-cell consists of a copper electrode in 1.00 M copper(II) nitrate, and the other consists of a copper electrode in a saturated solution of Cu2[Fe(CN)6].

Ferrocyanide, ([Fe(CN)6]4−), is a complex ion.

Answer 1

Step-by-step explanation:

Expression for
`K_(sp)` of the given reaction is as follows.

`K_(sp) = [Cu^(2+)]^(2)[Fe(CN)_(6)]`

Let us assume that the concentration of given species is "s". As the value of
`K_(sp)` is given as
`1.3 * 10^(-16)`.

`K_(sp) = [Cu^(2+)]^(2)[Fe(CN)_(6)]`

`1.3 * 10^(-16) = s^(2) * s`

`s^(3) = 1.3 * 10^(-16)`

s =
`3.19 * 10^(-6)`

Therefore, concentration of
`Cu^(2+)` will be calculated as follows.

`Cu^(2+)` = 2s

=
`2 * 3.19 * 10^(-6)`

=
`6.38 * 10^(-6)` M

Now, we will calculate the value of
`E_(cell)` as follows.

`E_(cell) = E^(o)_(cell) - (0.0591)/(2) * log (6.38 * 10^(-6))/(1)`

=
`0 - (0.0591)/(2) * log (6.38 * 10^(-6))/(1)`

= 0.1535 V

Thus, we can conclude that the potential of given cell is 0.1535 V.