Answer:
The voltage is [tex]E = 0.268V[/tex]
Explanation:
Since the main source of [tex]Na Cl[/tex] is the sea water then
The equation for the reaction [tex]Cl^{-}[/tex] in sea water would be
[tex]Cl^- ----->\frac{1}{2}Cl_2 + e^-[/tex]
that is the ion would form a molecule [tex]Cl_2[/tex] and then release an electron
For river water
[tex]\frac{1}{2}Cl_2 + e^- -----> Cl^-[/tex]
I.e the molecule would gain an electron and forms an ion
Now [tex]Cl^-[/tex] acts as reactant for seawater reaction and product for river water
The voltage of the concentration cell at initial stage is [tex]E_0 = 0[/tex]
Now at the voltage for the cell with [tex]Cl^-[/tex] concentration difference between sea water and river water is mathematical denoted as
[tex]E = E_0 - \frac{0.059}{n_e} log\frac{[Cl^-]_{product}}{[Cl^-]_{reactant}}[/tex]
Where [tex]n_e[/tex] is the number of electron = 1
So substituting values we have
[tex]E = 0 - \frac{0.059}{1} log\frac{1.0*10^{-3}}{35}[/tex]
Removing the negative sign
[tex]E = \frac{0.059}{1} log\frac{35}{1.0*10^{-3}}[/tex]
[tex]E = 0.268V[/tex]
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
[tex]Cl^- \rightarrow \frac{1}{2} Cl_2+e^-\\\\\frac{1}{2} Cl_2+e^- \rightarrow Cl^-[/tex]
Therefore,
It is concentration cell [tex]E^0 = 0[/tex]
[tex]E = E^0-\frac{0.059}{n} log\frac{[Cl^-]_{product}}{[Cl^-]_{rreactant}}[/tex]
[tex]Q = \frac{[Product]}{[Reactant]}[/tex]
[tex]E = Q - \frac{0.059}{1} log\frac{[Cl^-]_{riverwater}}{[Cl^-]_{seawater}}[/tex]
⇒[tex]0.059log\frac{[Cl^-]_{seawater}}{[Cl^-]_{riverwater}}[/tex]
[tex][Cl^-]_{seawater} = \frac{35}{35.5}[/tex]
[tex][Cl^-]_{riverwater}=\frac{10^-^3}{35.5}[/tex]
⇒ 0.059 log (35 × 10³)
E = 0.268V
Hence, the voltage (E) of a concentration cell constructed is 0.268V
