Question

Calculate the voltage (E) of a concentration cell constructed with the Cl- concentration difference between sea water and river water at 25ËC. Assume that the Cl- concentration (due to dissolved NaCl) of sea water is 35 g/L and that of river water is 1.0 mg/L.

Answer 1

the voltage (E) of a concentration cell constructed is 0.268V

Step-by-step explanation:

Chlorine concentration cell

Pt, Cl₂ / Cl⁻ // Cl⁻, Cl₂ , Pt

sea water River water

`Cl^- \rightarrow (1)/(2) Cl_2+e^-\\\\(1)/(2) Cl_2+e^- \rightarrow Cl^-`

Therefore,

It is concentration cell
`E^0 = 0`

`E = E^0-(0.059)/(n) log([Cl^-]_(product))/([Cl^-]_(rreactant))`

`Q = ([Product])/([Reactant])`

`E = Q - (0.059)/(1) log([Cl^-]_(riverwater))/([Cl^-]_(seawater))`

⇒
`0.059log([Cl^-]_(seawater))/([Cl^-]_(riverwater))`

`[Cl^-]_(seawater) = (35)/(35.5)`

`[Cl^-]_(riverwater)=(10^-^3)/(35.5)`

⇒ 0.059 log (35 × 10³)

E = 0.268V

Hence, the voltage (E) of a concentration cell constructed is 0.268V

Answer 2

The voltage is
`E = 0.268V`

Step-by-step explanation:

Since the main source of
`Na Cl` is the sea water then

The equation for the reaction
`Cl^(-)` in sea water would be

`Cl^- ----->(1)/(2)Cl_2 + e^-`

that is the ion would form a molecule
`Cl_2` and then release an electron

For river water

`(1)/(2)Cl_2 + e^- -----> Cl^-`

I.e the molecule would gain an electron and forms an ion

Now
`Cl^-` acts as reactant for seawater reaction and product for river water

The voltage of the concentration cell at initial stage is
`E_0 = 0`

Now at the voltage for the cell with
`Cl^-` concentration difference between sea water and river water is mathematical denoted as

`E = E_0 - (0.059)/(n_e) log([Cl^-]_(product))/([Cl^-]_(reactant))`

Where
`n_e` is the number of electron = 1

So substituting values we have

`E = 0 - (0.059)/(1) log(1.0*10^(-3))/(35)`

Removing the negative sign

`E = (0.059)/(1) log(35)/(1.0*10^(-3))`

`E = 0.268V`