Question

$E_{F{e}^{3+}/F{e}^{2+}}^o=0.771V$ and $E_{A{g}^{+}/Ag}^o=0.799V$
The minimum conc. of Ag${}^{+}$ that would remain unreduced by a standard $F{e}^{3+}/F{e}^{2+}$ electrode at equilibrium is:

• A. $3.35$ $mol$ $litre$${}^{-1}$
  
• B. $0.0335$ $mol$ $litre$${}^{-1}$
  
• C. $0.335$ $mol$ $litre$${}^{-1}$
• D. None of the above

Answer 1

Answer: (C)

$0.335$ $mol$ $litre$${}^{-1}$

Cathode: $A{g}^{+}+e^{-}\to Ag$
Anode: $F{e}^{2+}\to F{e}^{3+}+e^{-}$
Overall reaction: $A{g}^{+}+F{e}^{2+}\rightleftharpoons Ag+F{e}^{3+}$

The expression for the potential of the half electrode is-

$\Delta E=\Delta E^0-\frac{0.0592}{n}log\frac{\left[Ag\right]\left[F{e}^{2+}\right]}{\left[A{g}^{+}\right]\left[F{e}^{3+}\right]}$
At eqm. $\left[Ag\right]=1,\left[F{e}^{2+}\right]=\left[F{e}^{3+}\right];$ $\Delta E=0$, $\Delta E^o=E_{red}^o-E_{ox}^o=0.799-771$

Substitute values in the above expression.

$0=0.799-0.771-\frac{0.0592}{1}log\frac{1}{\left[A{g}^{+}\right]}$

$-log\left[A{g}^{+}\right]=\frac{0.799-0.771}{0.0592}=0.4729$

$\left[A{g}^{+}\right]=0.335M$.

Hence, the minimum conc. of Ag${}^{+}$ that would remain unreduced by a standard $F{e}^{3+}/F{e}^{2+}$ electrode at equilibrium is $0.335$ $mol$ $litre$${}^{-1}$.