Kd for dissociation of -AgNH32- into Ag- and NH3 is 6- 10-8- Calculate E- for the following half reaction- AgNH32-e- Ag-2NH3Given Ag-e- Ag- E-0-799 V

The correct option is B 0.373 V Ag→ Ag^++e^ #160; E_RP=0.799 V Ag(NH_3)_2^++e→ Ag+2 NH_3 #160; E_RP=? Ag(NH_3)_2^+⇌ Ag^++2 ...

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Electrochemical Cells, Galvanic Cells

Kd for dissoc...

Question

$K_{d}$ for dissociation of $[A g (N H_{3})_{2}]^{+}$ into $A g^{+}$ and $N H_{3}$ is $6 \times 10^{- 8}$ . Calculate $E^{\circ}$ for the following half reaction;
$A g (N H_{3})_{2}^{+} + e^{-} \to A g + 2 N H_{3}$
Given $A g^{+} + e^{-} \to A g, E^{\circ} = 0.799 V$

0.456 V

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0.373 V

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0.893 V

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0.593 V

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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})_{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

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}^{+} \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}

0.2 M

[A g {({N H}_{3})}_{2}] {N O}_{3}

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

K_{d i s s .} = \frac{[{A g}^{+}] {[{N H}_{3}]}^{2}}{[A g {({N H}_{3})}_{2}^{+}]} = 6.8 \times 10^{- 8}

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