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.

The voltage is

Explanation:

Since the main source of is the sea water then

   The equation for the reaction    in sea water would be

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

For river water

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

Now   acts as reactant for seawater reaction and product for river water

    The voltage of the concentration cell at initial stage is

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

Where is the number of electron = 1

So substituting values we have

   Removing the negative sign

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

Explanation:

Chlorine concentration cell

   Pt, Cl₂ / Cl⁻    //    Cl⁻, Cl₂ , Pt

   sea water             River water

Therefore,

It is concentration cell

⇒

⇒ 0.059 log (35 × 10³)

E = 0.268V

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

answer; i believe the correct answer is (c) /// is a homogeneous mixture of gaseous substance nitrogen,oxygen and smaller amounts of other substancesmixtures can be either ( homogeneous or heterogeneous)plz check first good luck