Consider the cell Ag s - AgBr s Br - aq AgCl s - Cl - aq - Ag s 25- C - The solubility product constants of AgBr and AgCl are respectively 5 - 10-13 and1 - 10-10- For what ratio of the concentration of Br -and Cl -ions would the emf of the cell be zero-A- 1- 100B- 1- 50C- 1- 25D- 1- 200

The correct option is D 1 : 200[Ag^+]_RHS = 1 × 10^ 10/[Cl^ ]; [Ag^+]_LHS = 5 × 10^ 13/[Br^ ] E_Cell = 0.0591/1 log [Ag^+]_RHS/[Ag^+]_LHS 0 = 0.591 × log 1 × 1 ...

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Byju's Answer

Standard XII

Chemistry

Dipole Moment

Consider the ...

Question

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

1 : 200

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1 : 100

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1 : 50

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1 : 25

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Solution

The correct option is A 1 : 200
$[A g^{+}]_{R H S} = \frac{1 \times 10^{- 10}}{[C l^{-}]}; [A g^{+}]_{L H S} = \frac{5 \times 10^{- 13}}{[B r^{-}]}$
$E_{C e l l} = \frac{0.0591}{1} l o g \frac{[A g^{+}]_{R H S}}{[A g^{+}]_{L H S}}$
$0 = 0.591 \times l o g \frac{\frac{1 \times 10^{- 10}}{[C l^{-}]}}{\frac{5 \times 10^{- 13}}{[B r^{-}]}} \Rightarrow l o g \frac{\frac{1 \times 10^{- 10}}{[C l^{-}]}}{\frac{5 \times 10^{- 13}}{[B r^{-}]}} = 0$
$\frac{\frac{1 \times 10^{- 10}}{[C l^{-}]}}{\frac{5 \times 10^{- 13}}{[B r^{-}]}} = 1 \Rightarrow \frac{5 \times 10^{- 13}}{[B r^{-}]} = \frac{1 \times 10^{- 10}}{[C l^{-}]}$
$\frac{[B r^{-}]}{[C l^{-}]} = \frac{5 \times 10^{- 13}}{1 \times 10^{- 10}} = 0.005 = \frac{1}{200}$

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Consider the cell Ag(s)|AgBr(s)||Br−(aq)||AgCl(s)|Cl−(aq)|Ag(s)25∘C.The solubility product constants of AgBr and AgCl are respectively 5×10−13 and 1×10−10. For what ratio of the concentration of Br− and Cl− ions would the emf of the cell be zero?