Consider the cell Ag - AgBr s Br- - AgCls Cl- - Ag at 25oC- The solubility product of AgCl and AgBr are 1 - 10-10 and 5 - 10-13 respectively- For what ratio of concentration of Br- and Cl- ions would the emf of cell be zero- give ans in1-litre

The expression for the cell potential is E=E^0 0.0592nlog[Ag^+]_L[Ag^+]_R Substitute values in the above expression 0=0 0.05921log[Ag^+]_L[Ag^+]_R Henc ...

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Standard XII

Chemistry

Electrode Potential

Consider the ...

Question

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

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Solution

The expression for the cell potential is
$E = E^{0} - \frac{0.0592}{n} l o g \frac{[A g^{+}]_{L}}{[A g^{+}]_{R}}$
Substitute values in the above expression
$0 = 0 - \frac{0.0592}{1} l o g \frac{[A g^{+}]_{L}}{[A g^{+}]_{R}}$
Hence, $\frac{[A g^{+}]_{L}}{[A g^{+}]_{R}} = 1$
or $[A g^{+}]_{L} = [A g^{+}]_{R}$
Hence, $\frac{K_{s p, A g B r}}{[B r^{-}]} = \frac{K_{s p, A g C l}}{[C l^{-}]}$
$\frac{[C l^{-}]}{[B r^{-}]} = \frac{K_{s p, A g C l}}{K_{s p, A g B r}} = \frac{1 \times 10^{- 10}}{5 \times 10^{- 3}} = 200$

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Consider the cell Ag|AgBr(s)Br−||AgCl(s)Cl−|Ag at 25oC. The solubility product of AgCl and AgBr are 1×10−10 and 5×10−13 respectively. For what ratio of concentration of Br− and Cl− ions would the emf of cell be zero? give ans in(1/litre)

The expression for the cell potential is E=E0−0.0592nlog[Ag+]L[Ag+]RSubstitute values in the above expression0=0−0.05921log[Ag+]L[Ag+]RHence, [Ag+]L[Ag+]R=1or [Ag+]L=[Ag+]RHence, Ksp,AgBr[Br−]=Ksp,AgCl[Cl−][Cl−][Br−]=Ksp,AgClKsp,AgBr=1×10−105×10−3=200