Chemistry - Electrochemistry Question with Solution | TestHub

The Nernst equation is used to calculate the cell potential ($E_{cell}$) under non-standard conditions:

$$E_{cell}=E_{cell}^0-\frac{RT}{nF}\ln Q$$

Given $\frac{RT}{F}=0.06$ at 25ºC, and $n=1$ for all Ag/Ag+ electrode reactions, the equation becomes:

$$E_{cell}=E_{cell}^0-0.06\log Q$$

For concentration cells, $E_{cell}^0=0$.

Equilibrium constants provided:

1. $\mathrm{AgCl}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Cl}}^{-}{K}_{sp}(\mathrm{AgCl})=10^{-10}$

2. $\mathrm{AgBr}\rightleftharpoons {\mathrm{Ag}}^{+}+{\mathrm{Br}}^{-}{K}_{sp}(\mathrm{AgBr})=10^{-12}$

3. $\mathrm{Ag}({\mathrm{NH}}_3{)}^{2+}\rightleftharpoons {\mathrm{Ag}}^{+}+2{\mathrm{NH}}_3{K}_{inst}(\mathrm{Ag}({\mathrm{NH}}_3{)}^{2+})=10^{-8}$

Let's analyze each cell:

(P) ${\mathrm{Ag}}_{(\mathrm{s})}|{\mathrm{AgBr}}_{(\mathrm{s})}|{\mathrm{Br}}^{-}{}_{(\mathrm{aq})}||{\mathrm{Cl}}^{-}{}_{(\mathrm{aq})}|{\mathrm{AgCl}}_{(\mathrm{s})}|{\mathrm{Ag}}_{(\mathrm{s})}$

This is a concentration cell with different anion concentrations.

Anode (oxidation): ${\mathrm{Ag}}_{(\mathrm{s})}+{\mathrm{Br}}^{-}{}_{(\mathrm{aq})}\to {\mathrm{AgBr}}_{(\mathrm{s})}+{\mathrm{e}}^{-}$

Cathode (reduction): ${\mathrm{AgCl}}_{(\mathrm{s})}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{(\mathrm{s})}+{\mathrm{Cl}}^{-}{}_{(\mathrm{aq})}$

Overall reaction: ${\mathrm{AgBr}}_{(\mathrm{s})}+{\mathrm{Cl}}^{-}{}_{(\mathrm{aq})}\to {\mathrm{AgCl}}_{(\mathrm{s})}+{\mathrm{Br}}^{-}{}_{(\mathrm{aq})}$

The cell potential is given by:

$E_{cell}=E_{AgCl/Ag}^0-E_{AgBr/Ag}^0$

We know $E_{AgX/Ag}^0=E_{A{g}^{+}/Ag}^0+0.06\log K_{sp}(AgX)$.

$E_{AgCl/Ag}^0=E_{A{g}^{+}/Ag}^0+0.06\log (10^{-10})$

$E_{AgBr/Ag}^0=E_{A{g}^{+}/Ag}^0+0.06\log (10^{-12})$

$E_{cell}=(E_{A{g}^{+}/Ag}^0+0.06\log (10^{-10}))-(E_{A{g}^{+}/Ag}^0+0.06\log (10^{-12}))$

$E_{cell}=0.06(-10-(-12))=0.06\times 2=0.12$ V.

So, (P) matches (3) 0.12 volt.

(Q) ${\mathrm{Ag}}_{(\mathrm{s})}|{\mathrm{AgCl}}_{(\mathrm{s})}|{\mathrm{Cl}}^{-}{}_{(\mathrm{aq})}||{\mathrm{Ag}}^{+}{}_{(\mathrm{aq})}|{\mathrm{Ag}}_{(\mathrm{s})}$

Anode: ${\mathrm{Ag}}_{(\mathrm{s})}+{\mathrm{Cl}}^{-}{}_{(\mathrm{aq})}\to {\mathrm{AgCl}}_{(\mathrm{s})}+{\mathrm{e}}^{-}$

Cathode: ${\mathrm{Ag}}^{+}{}_{(\mathrm{aq})}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{(\mathrm{s})}$

The standard potential for the anode reaction is $E_{Ag/AgCl,C{l}^{-}}^0=E_{A{g}^{+}/Ag}^0+0.06\log K_{sp}(\mathrm{AgCl})$.

The standard potential for the cathode reaction is $E_{A{g}^{+}/Ag}^0$.

$E_{cell}^0=E_{cathode}^0-E_{anode}^0=E_{A{g}^{+}/Ag}^0-(E_{A{g}^{+}/Ag}^0+0.06\log K_{sp}(\mathrm{AgCl}))$

$E_{cell}^0=-0.06\log (10^{-10})=-0.06\times (-10)=0.6$ V.

So, (Q) matches (4) 0.6 volt.

(R) ${\mathrm{Ag}}_{(\mathrm{s})}|{\mathrm{AgBr}}_{(\mathrm{s})}|{\mathrm{Br}}^{-}{}_{(\mathrm{aq})}||\mathrm{Ag}({\mathrm{NH}}_3{)}_2{}^{+}{}_{(\mathrm{aq})}|{\mathrm{Ag}}_{(\mathrm{s})}$

Anode: ${\mathrm{Ag}}_{(\mathrm{s})}+{\mathrm{Br}}^{-}{}_{(\mathrm{aq})}\to {\mathrm{AgBr}}_{(\mathrm{s})}+{\mathrm{e}}^{-}$

Cathode: $\mathrm{Ag}({\mathrm{NH}}_3{)}_2{}^{+}{}_{(\mathrm{aq})}+{\mathrm{e}}^{-}\to {\mathrm{Ag}}_{(\mathrm{s})}+2{\mathrm{NH}}_3$

The standard potential for the anode reaction is $E_{Ag/AgBr,B{r}^{-}}^0=E_{A{g}^{+}/Ag}^0+0.06\log K_{sp}(\mathrm{AgBr})$.

The standard potential for the cathode reaction is $E_{Ag(N{H}_3{)}_2^{+}/Ag}^0=E_{A{g}^{+}/Ag}^0+0.06\log K_{inst}(\mathrm{Ag}({\mathrm{NH}}_3{)}^{2+})$.

$E_{cell}^0=E_{Ag(N{H}_3{)}_2^{+}/Ag}^0-E_{Ag/AgBr,B{r}^{-}}^0$

$E_{cell}^0=(E_{A{g}^{+}/Ag}^0+0.06\log K_{inst}(\mathrm{Ag}({\mathrm{NH}}_3{)}^{2+}))-(E_{A{g}^{+}/Ag}^0+0.06\log K_{sp}(\mathrm{AgBr}))$

$E_{cell}^0=0.06(\log (10^{-8})-\log (10^{-12}))=0.06(-8-(-12))=0.06\times 4=0.24$ V.

So, (R) matches (2) 0.24 volt.

(S) $\mathrm{Pt},{\mathrm{H}}_{2(\mathrm{g})}|{\mathrm{H}}^{+}{}_{(\mathrm{aq})}||{\mathrm{H}}_{2(\mathrm{g})},\mathrm{Pt}$

This is a hydrogen concentration cell.

Anode: ${\mathrm{H}}_{2(\mathrm{g})}\to 2{\mathrm{H}}^{+}{}_{(\mathrm{aq})}+2{\mathrm{e}}^{-}$

Cathode: $2{\mathrm{H}}^{+}{}_{(\mathrm{aq})}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_{2(\mathrm{g})}$

Since the species and their concentrations/pressures are identical on both sides, the cell potential is 0.

So, (S) matches (1) 0 volt.

Matching:

P $\to$ 3

Q $\to$ 4

R $\to$ 2

S $\to$ 1

Final Answer: The final answer is A