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

Answer:

The voltage is $E = 0.268V$

Explanation:

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

   The equation for the reaction  $Cl^{-}$  in sea water would be

              $Cl^- ----->\frac{1}{2}Cl_2 + e^-$

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

For river water

               $\frac{1}{2}Cl_2 + e^- -----> Cl^-$

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

                $E = E_0 - \frac{0.059}{n_e} log\frac{[Cl^-]_{product}}{[Cl^-]_{reactant}}$

Where $n_e$ is the number of electron = 1

So substituting values we have

              $E = 0 - \frac{0.059}{1} log\frac{1.0*10^{-3}}{35}$

   Removing the negative sign

              $E = \frac{0.059}{1} log\frac{35}{1.0*10^{-3}}$

                  $E = 0.268V$

                 

                   

             

Answer 2

Answer:

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

$Cl^- \rightarrow \frac{1}{2} Cl_2+e^-\\\\\frac{1}{2} Cl_2+e^- \rightarrow Cl^-$

Therefore,

It is concentration cell $E^0 = 0$

$E = E^0-\frac{0.059}{n} log\frac{[Cl^-]_{product}}{[Cl^-]_{rreactant}}$

$Q = \frac{[Product]}{[Reactant]}$

$E = Q - \frac{0.059}{1} log\frac{[Cl^-]_{riverwater}}{[Cl^-]_{seawater}}$

⇒$0.059log\frac{[Cl^-]_{seawater}}{[Cl^-]_{riverwater}}$

$[Cl^-]_{seawater} = \frac{35}{35.5}$

$[Cl^-]_{riverwater}=\frac{10^-^3}{35.5}$

⇒ 0.059 log (35 × 10³)

E = 0.268V

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