Dissociation constant for AgNH3-2 into Ag- and NH3 is 6 - 10-14- Eo for the half cell reaction is Ag NH32- - e - Ag - NH3Given Ag- - e - Ag- Eo - 0-799 V The answer is 0-0mn then m-n -

nbsp; nbsp; nbsp; nbsp; nbsp; Ag nbsp; → Ag^+ nbsp;+ e ; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbs ...

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Concentration Terms - An Introduction (w/w etc)

Dissociation ...

Question

Dissociation constant for $A g (N H_{3})_{2}^{+}$ into $A g^{+}$ and $N H_{3}$ is $6 \times 10^{- 14}$ , $E^{o}$ for the half cell reaction is
$A g (N H_{3})_{2}^{+} + e \to A g + N H_{3}$
Given $A g^{+} + e \to A g; E^{o} = 0.799 V$ The answer is 0.0mn then m+n =

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

6 \times 10^{14}

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

E^{o} A g^{+} / A g = 0.799 V

E^{o}

[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

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}

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}]^{+}

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