Equilibrium constant for the dissociation of Ag ( NH _ 3 )_ 2 ^ + into Ag + and NH 3 is 6…

Chemistry · JEE Main · NTA ExamsRedox Reactions and Electrochemistry

Equilibrium constant for the dissociation of \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) into Ag+  and NH3 is 6 x 10-8.  Calculate Eo for the following half reaction.

\(\left[\mathrm{Ag}\left[\mathrm{NH}_{3}\right]_{2}\right]^{+}+\mathbf{e}^{-} \longrightarrow \mathrm{Ag}(5)+2 \mathrm{NH}_{3}\)

\(\left[E_{\mathrm{Ag}^{+} / \mathrm{Ag}^{+}}=0.799 \mathrm{~V}, \log 6=0.7781\right]\)

  1. 0.372
  2. 0.799
  3. 0.7781
  4. 0.6

Answer

(A) 0.372

Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.

Related practice questions

More Redox Reactions and Electrochemistry questions · Browse all practice questions