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Concentration Terms - An Introduction (w/w etc)

Given that E0...

Question

Given that $E_{C l_{2} / C l^{-}}^{0} = 1.35 V$ and $K_{s p} (A g C l) = 10^{- 10}$ at $25^{o} C$ .

$E^{0}$ corresponding to the electrode reaction is
$\frac{1}{2} C l_{2} (g) + A g^{+} (a q) + e^{-} \to A g C l (s)$ is

0.75 V

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1.05 V

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2.65 V

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1.94 V

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Solution

The correct option is D $1.94 V$
For sparingly soluble $A g C l$
$A g C l (s) \to A g^{+} (a q) + C l^{-} (a q) . . . (e q 1)$
$K_{s p} (A g C l) = 10^{- 10}$
$Δ G_{1} = - R T ln K_{s p}$
For :
$\frac{1}{2} C l_{2} (g) + e^{-} \to C l^{-} (a q) . . . . . . (e q 2); E_{2}^{0} = 1.35 V$
$Δ G_{2} = - n F E_{2}^{0}$
Reversing eqn 1 and adding with equation 2, we get the desired equation
$\frac{1}{2} C l_{2} (g) + A g^{+} (a q) + e^{-} \to A g C l (s) . . . (e q 3)$
$Δ G_{3} = - n F E_{3}^{0} = Δ G_{2} - Δ G_{1}$
$- n F E_{3}^{0} = - n F E_{2}^{0} - (- R T ln K_{s p})$
$E_{3}^{0} = E_{2}^{0} - \frac{R T}{F} ln 10^{- 10} E_{3}^{0} = 1.35 + (0.059 \times 10) = 1.94 V$

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Similar questions

A g^{+} | A g

0.80 v o l t

C l^{-} | A g C l | A g

25^{\circ} C

K_{s p} (A g C l) = 1.8 \times 10^{- 10}

A g^{+} (a q) + e^{-} \to A g (s)

E^{0} = + 0.799 V

A g C l (s) + e^{-} \to A g (s) + C l^{-} (a q)

E^{0} = + 0.222 V

1.743 \times 10^{- 10}

A g C l_{(s)} + e^{-} \to A g_{(s)} + C l_{(a q)}^{-}

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A g_{(a q)}^{+} + e^{-} \to A g_{(s)}; E^{⊖} = 0.80

A g C l

A g | A g C l (s), K C l (0.2 M) | | K B r (0.001 M), A g B r (s) | A g

25^{\circ} C

K_{s p A g C l} = 2.8 \times 10^{- 10}; K_{s p A g B r} = 3.3 \times 10^{- 13}

A g | A g C l (s), K C l (0.2 M) | | K B r (0.001 M), A g B r (s) | A g

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[K_{s p} (A g C l) = 2.8 \times 10^{- 10}, K_{s p} (A g B r) = 3.3 \times 10^{- 13}]

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