Question

Consider the cell ${\text{Ag|AgBr(s)B}}{{\text{r}}^{\text{ - }}}{\text{||AgCl(s)C}}{{\text{l}}^{\text{ - }}}{\text{|Ag}}$at ${\text{25}}^\circ {\text{C}}$. The solubility product of ${\text{AgCl}}$and ${\text{AgBr}}$are ${\text{1}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}{\text{and 5}} \times {\text{1}}{{\text{0}}^{{\text{ - 13}}}}$respectively. For what ratio of concentration of ${\text{B}}{{\text{r}}^{\text{ - }}}$and ${\text{C}}{{\text{l}}^{\text{ - }}}$would the emf of cell be zero? Give and in (1/litre)

Answer 1

Use the Nernst equation and calculate the ratio of the concentration of ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}}/{{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$for zero emf of the cell. Using this ratio of concentration of ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}}/{{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$ and solubility product values calculate the ratio of the concentration of ${\text{B}}{{\text{r}}^{\text{ - }}}$and ${\text{C}}{{\text{l}}^{\text{ - }}}$.

Formulas Used:
Nernst Equation:
${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{0}}} + \dfrac{{0.0592}}{{\text{n}}}\log \dfrac{{{\text{[Anode]}}}}{{{\text{[Cathode]}}}}$

${K_{sp}}{\text{(AgCl)}} = {\text{[A}}{{\text{g}}^{\text{ + }}}][{\text{C}}{{\text{l}}^{\text{ - }}}]$
${K_{sp}}{\text{(AgBr)}} = {\text{[A}}{{\text{g}}^{\text{ + }}}][B{r^{\text{ - }}}]$

Complete step-by-step answer: Cell given to us is ${\text{Ag|AgBr(s)B}}{{\text{r}}^{\text{ - }}}{\text{||AgCl(s)C}}{{\text{l}}^{\text{ - }}}{\text{|Ag}}$

As per the cell notation, double lines indicate the salt bridge which separates the two half cell reactions. Half cell reaction on the left side of the salt bridge is the anodic reaction. Half cell reaction on the right side of the salt bridge is a cathodic reaction. Oxidation takes place at the anode while reduction takes place at the cathode.
Use the Nernst equation and calculate the ratio of the concentration of ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}}/{{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$for zero emf of the cell.

Nernst Equation:

${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{0}}} + \dfrac{{0.0592}}{{\text{n}}}\log \dfrac{{{\text{[Anode]}}}}{{{\text{[Cathode]}}}}$

Where,

n= number of electrons transfer

For the given cell there is a transfer of 1 electron.

${\text{ }}{{\text{E}}^{\text{0}}}$ cell for the given reaction is zero as the same species is getting oxidized and reduced.

${{\text{E}}_{{\text{cell}}}}{\text{ = }}{{\text{E}}^{\text{0}}} + \dfrac{{0.0592}}{{\text{n}}}\log \dfrac{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{anode}}}}}}{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{cathode}}}}}}$

Now, substitute zero for ${{\text{E}}_{{\text{cell}}}}$ , zero for ${\text{ }}{{\text{E}}^{\text{0}}}$ and 1 for the number of electron transfer and calculate the ratio ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}}/{{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$

$\Rightarrow {\text{0 = 0}} + \dfrac{{0.0592}}{1}\log \dfrac{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{anode}}}}}}{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{cathode}}}}}}$

$\Rightarrow \dfrac{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{anode}}}}}}{{{{{\text{[A}}{{\text{g}}^{\text{ + }}}]}_{{\text{cathode}}}}}} = 1$

Now, using the solubility product of ${\text{AgCl}}$and ${\text{AgBr}}$ and ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}}/{{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$ ratio calculates the ratio of the concentration of ${\text{B}}{{\text{r}}^{\text{ - }}}$and ${\text{C}}{{\text{l}}^{\text{ - }}}$ as follows:

${K_{sp}}{\text{(AgBr)}} = {{\text{[A}}{{\text{g}}^{\text{ + }}}]_{anode}}[B{r^{\text{ - }}}]$

$\Rightarrow {{\text{[A}}{{\text{g}}^{\text{ + }}}]_{anode}} = \dfrac{{{K_{sp}}{\text{(AgBr)}}}}{{[B{r^{\text{ - }}}]}}$

$\Rightarrow {{\text{[A}}{{\text{g}}^{\text{ + }}}]_{cathode}} = \dfrac{{{K_{sp}}{\text{(AgCl)}}}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}}$

As ${{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{anode}}}} = {{\text{[A}}{{\text{g}}^{\text{ + }}}]_{{\text{cathode}}}}$

So, $\dfrac{{{K_{sp}}{\text{(AgBr)}}}}{{[B{r^{\text{ - }}}]}} = \dfrac{{{K_{sp}}{\text{(AgCl)}}}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}}$

$\dfrac{{{K_{sp}}{\text{(AgBr)}}}}{{{K_{sp}}{\text{(AgC}})}} = \dfrac{{[B{r^{\text{ - }}}]}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}}$

Now, substitute ${\text{1}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}$ for solubility product of ${\text{AgCl}}$ and ${\text{5}} \times {\text{1}}{{\text{0}}^{{\text{ - 13}}}}$ for solubility product of${\text{AgBr}}$ and calculate the ratio of the concentration of ${\text{B}}{{\text{r}}^{\text{ - }}}$and.

$\dfrac{{[B{r^{\text{ - }}}]}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}} = \dfrac{{{\text{5}} \times {\text{1}}{{\text{0}}^{{\text{ - 13}}}}}}{{{\text{1}} \times {\text{1}}{{\text{0}}^{{\text{ - 10}}}}}}$

$\Rightarrow \dfrac{{[B{r^{\text{ - }}}]}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}} = 0.005$
$\Rightarrow \dfrac{{[B{r^{\text{ - }}}]}}{{[{\text{C}}{{\text{l}}^{\text{ - }}}]}} = \dfrac{1}{{200}}$
Thus, at a ratio of 1/200 for the concentration of ${\text{B}}{{\text{r}}^{\text{ - }}}$and ${\text{C}}{{\text{l}}^{\text{ - }}}$the emf of the cell would be zero.

Note: Solubility values of ${\text{AgCl}}$and ${\text{AgBr}}$ are very low which indicate that these salt are sparingly soluble. The standard electrode potential of the cell is the potential difference between the standard electrode potential of the right-hand cell (cathode) minus the standard reduction potential of the left-hand cell (anode).