Dissociation constant for- AgNH3-2 - Ag- - NH3 is 6 - 1014- AgNH32- - e - Ag - 2 NH3 if Eo Ag-Ag - 0-799 V Eo for the half reaction is multiply answer with 100 -

Dissociation constant = 6*10^14 #160;and, E^0_cell = 0.0591 * log K we get E^0_cell = 0.873so, E^0 = 0.873 0.799 = 0.074thus , we hac ...

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Heat of Reaction

Dissociation ...

Question

Dissociation constant for:
$A g (N H_{3})_{2}^{+} \to A g^{+} + N H_{3}$ is $6 \times 10^{14}$ .
$A g (N H_{3})_{2}^{+} + e \to A g + 2 N H_{3}$ if $E^{o} A g^{+} / A g = 0.799 V$
$E^{o}$ for the half reaction is (multiply answer with 100) :

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A g (N H_{3})_{2}^{+}

A g^{+}

N H_{3}

6 \times 10^{- 14}

E^{o}

A g (N H_{3})_{2}^{+} + e \to A g + N H_{3}

A g^{+} + e \to A g; E^{o} = 0.799 V

[A g (N H_{3})_{2}]^{+} ⇌ A g^{+} + 2 N H_{3}

6 \times 10^{- 14} .

E^{o}

y \times 10^{- 3} e V

[A g (N H_{3})_{2}]^{+} + e^{-} \to A g + 2 N H_{3}

A g^{+} + e^{-} \to A g; E^{0} = 0.799 V

K_{d}

[A g (N H_{3})_{2}]^{+}

A g^{+}

N H_{3}

6 \times 10^{- 8}

E^{\circ}

A g (N H_{3})_{2}^{+} + e^{-} \to A g + 2 N H_{3}

A g^{+} + e^{-} \to A g, E^{\circ} = 0.799 V

E^{\circ}

298 K

A g (N H_{3})_{2}^{+} + e^{-} \to A g + 2 N H_{3}

A g^{+} + e^{-} \to A g; E_{A g^{+} / A g}^{\circ} = 0.80 V

A g (N H_{3})_{2}^{+} ⇌ A g^{+} + 2 N H_{3}; K = 6 \times 10^{- 8}

A g^{+} + N H_{3} ⇌ A g (N H_{3})^{+}

K_{1} = 3.5 \times 10^{- 3}

A g (N H_{3}) + N H_{3} ⇌ [A g (N H_{3})_{2}]

K_{2} = 1.7 \times 10^{- 3}

[A g (N H_{3})_{2}]^{+}

A g^{+}

N H_{3}

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