$C u I (s) + e^{-} \to C u_{(s)} + I_{(a q)}^{-}$ ; $E^{\circ} = - 0.188 V$
$C u_{(a q)}^{2 +} + I^{-} + e^{-} \to C u I_{(s)} E^{\circ} = 0.868 V$ ;
$K_{s p} (C u I) = 10^{- 12} M^{2}$ and $\frac{298 R l n 10}{F} = 0.059$
If the equilibrium constant of the disproportionation reaction of $C u^{+}$ is $K_{c}$ i.e.
$2 C u^{+} (a q)_{c} C u (s) + C u^{2 +} (a q)$ ; then find the value of ( $0.59 \times l o g K_{c}$ )

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Solution

Given,
$C u I (s) ⇌ C u^{+} + I^{-} K_{s p} = 10^{- 12} M^{2}$
So when $[I^{-}] = 1 M$ , then $[C u^{+}] = 10^{- 12} M$
Again,
$C u I_{(s)} + e \to C u_{(s)} + I_{(a q)}^{-} E^{\circ} = - 0.188 V$ ;
This is equivalent to
$C u_{a q}^{+} 10^{- 12} + e^{-} \to C u_{(s)} E = - 0.188 V$
So,
$E = E_{1}^{\circ} - \frac{0.059}{1} l o g \frac{1}{10^{- 12}}$ $\Rightarrow - 0.188 = E_{1}^{\circ} - \frac{0.059}{1} l o g \frac{1}{10^{- 12}}$
$\Rightarrow E_{1}^{\circ} = 0.52 V$
Again,
$C u_{(a q)}^{2 +} + I^{-} + e^{-} \to C u I_{(s)} E^{\circ} = 0.868 V$
This is equivalent to
$C u_{(a q)}^{2 +} 1 M + e^{-} \to C u_{(a q)}^{+} 10^{- 12} M E = 0.868 V$
So,
$E = E_{2}^{\circ} - \frac{0.059}{1} l o g \frac{10^{- 12}}{1}$ $\Rightarrow 0.868 = E_{2}^{\circ} - \frac{0.059}{1} l o g \frac{10^{- 12}}{1}$
$\Rightarrow E_{2}^{\circ} = 0.16 V$
Now for disproportionation of $C u^{+}$ i.e.
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$ $E_{c e l l}^{\circ} = E_{1}^{\circ} - E_{2}^{\circ} = 0.52 - 0.16 = 0.36 V$
So For equilibrium $2 C u^{+} (a q)_{c} C u (s) + C u^{2 +} (a q)$
$E_{c e l l}^{\circ} = \frac{0.059}{1} l o g K_{c} \Rightarrow 0.59 l o g K_{c} = 3.6$

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Similar questions

Q. Given

C u I_{(s)} + e^{-} ⟶ C u_{(s)} + I_{(a q)}^{-}; E^{\circ} = - 0.188 V . . . (i) C u_{(a q)}^{2 +} + I^{-} + e^{-} ⟶ C u I_{(s)}; E^{\circ} = 0.868 V . . . (i i) K_{s p} (C u I) = 10^{- 12} M^{2} . . . (i i i)

\frac{298 R l n 10}{F} = 0.059

List -1 contains half cells and List-2 contains emf values of half cells

\begin{matrix} List - I & List - II (P) & C u_{(a q)}^{+} + e^{-} ⟶ C u_{(s)}; E_{1}^{0} = & (1) & 0.34 V (Q) & C u_{(a q)}^{2 +} + e^{-} ⟶ C u_{(a q)}^{+}; E_{2}^{0} = & (2) & 0.16 V (R) & C u_{(a q)}^{2 +} + 2 e^{-} ⟶ C u_{(s)}; E_{3}^{0} = & (3) & 0.52 V (S) & C u_{(1 M)}^{2 +} + I_{(10^{- 12} M)}^{-} + e^{-} ⟶ C u I (s); E_{4} = & (4) & (E_{1}^{0} - E_{3}^{0} - 0.02) V (5) & 2 \times E_{2}^{0} + 0.02 \end{matrix}

Match the emf of cell in LIST-I with one or more values in LIST-II and choose the correct option.

Given the half cell reactions:
$C u^{+} (a q) + e^{-} \to C u (s); E^{0} = + 0.52 V C u^{2 +} (a q) + e^{-} \to C u^{+} (a q); E^{0} = + 0.165 V$
The equilibrium constant for the disproportionation reaction at 298 K is:
$2 C u^{+} (a q) \to C u (s) + C u^{2 +} (a q)$