Question

The cell potential of the following electrochemical cell depends on the gold concentration in the cathode half-cell: Pt(s)|H2(g,1atm)|H+(aq,1.0M)|Au3+(aq,?M)|Au(s). What is the concentration of Au3+ in the solution if Ecell is 1.27 V ? Express your answer using two significant figures.

Answer 1

Step-by-step explanation:

The standard reduction potential
`(E_(cell))` is given as 1.27 V.

Concentration of
`H^(+)` = 1.0 M

Hence, reduction reaction is as follows.

`Au^(3+)(aq) + 3e^(-) \rightarrow 2Au(s)`

Oxidation reaction is as follows.

`H_(2)(g) \rightarrow H^(+)(aq) + e^(-)`

Therefore, overall net chemical equation will be as follows.

`2Au^(3+)(aq) + 3H_(2)(g) \rightarrow 2Au(s) + 6H^(+)(aq)`

As, its is known that standard electrode potential (
`E^(o)_(cell)`) for hydrogen is equal to zero.

And, standard electrode potential for
`Au^(3+)` is
`E_(o)` equal 1.50 V.

Hence, formula to calculate standard cell potential is as follows.

`E^(o)_(cell)` =
`(E_(o))_(reduction)` -
`(E_(o))_(oxidation)`

= 1.50 V - 0 V

= 1.50 V

Therefore, concentration of
`Au^(3+)` is calculated as follows.

`E_(cell) = E^(o)_(cell) - (0.059)/(2)log ([H^(+)]^(6))/([Au^(3+)]^(2))`

Hence, substitute the given values as follows.

`E_(cell) = E^(o)_(cell) - (0.059)/(2)log ([H^(+)]^(6))/([Au^(3+)]^(2))`

1.27 = 1.50 - \frac{0.059}{6}log \frac{(1)^{6}}{[Au^{3+}]^{2}}[/tex]

`log(1)/([Au^(3+)]^(2))` = 23

Taking antilog on both the sides, the above equation will be as follows.

`(1)/([Au^(3+)]^(2))` =
`10^(23)`

=
`1 * 10^(23)`

`(1)/([Au^(3+)]^(2))` =
`3.16 * 10^(11)`

`[Au^(3+)]` =
`3.16 * 10^(-12)` M

Hence, we can conclude that the concentration for
`[Au^(3+)]` is
`3.16 * 10^(-12)` M.