ElectrochemistryHard
Question
The standard reduction potential for the reactions: Ag+ + e− → Ag and Ag(NH3)2+ + e− → Ag + 2NH3 are +0.79 V and +0.37 V, respectively. From these values and the Nernst equation, what should be Kf for the Ag(NH3)2+ ion? [Given: 2.303 RT/F = 0.06]
Options
A.1.0 × 10−7
B.1.0 × 107
C.2.15 × 1019
D.4.64 × 10−20
Solution
$Ag^{+} + 2NH_{3} \rightleftharpoons Ag(NH)_{3}^{+};E^{o} = 0.79 - 0.37 = 0.42\text{ V}$
Now, $E^{o} - \frac{0.06}{n}.\log K_{eq} \Rightarrow 0.42 = \frac{0.06}{1}.\log K_{f} $$$\Rightarrow K_{f} = 10^{7}$$
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