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Upon inspection of the standard electrode potentials for the half reactions for both $\ce{Cu^2+}$/$\ce{Cu+}$ and $\ce{I2}$/$\ce{2I-}$, it becomes apparent it should be the iodide ions that act as the oxidising agent, as the standard electrode potential for the iodine half-cell is much higher. $$\begin{alignat}{2} \ce{Cu^2+ + e- \;&<=> Cu+}\quad &&E_\text{cell} = +0.159\ \mathrm{V}\\ \ce{I2 + 2e- \;&<=> 2I-}\quad &&E_\text{cell} = +0.54\ \mathrm{V}\\ \end{alignat}$$

To explain this, many websites that I’ve looked at suggest it’s the weak solubility of copper(I) iodide, increasing the $\ce{Cu^2+/Cu+}$ potential to around $+0.88\ \mathrm{V}$.

Such websites include:

What I’d like to know, is how this value of $+0.88\ \mathrm{V}$ has been determined, and a little of the theory behind as to why it has increased from $+0.159\ \mathrm{V}$.

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Let's consider the following redox couples:

$\ce{Cu^{2+} + e^- ->Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$

$\ce{Cu^{2+} + e^- + I^- ->CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$

As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, the two redox couples $\ce{Cu^{2+}/CuI}$ and $\ce{Cu^{2+}/Cu^+}$ are in equilibrium and that means: $$ E_1(\ce{Cu^{2+}/Cu^+})=E_2(\ce{Cu^{2+}/CuI})$$ Using Nernst equation for each couple:$$ E_1(\ce{Cu^{2+}/Cu^+})=E^0_1(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}]}}$$ $$ E_2(\ce{Cu^{2+}/CuI})=E^0_2(\ce{Cu^{2+}/CuI})+\frac{RT}{F}\ln{[\ce{I^-}][\ce{Cu^2+}]}$$ We can write then: $$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{[\ce{Cu^2+}]}{[\ce{Cu^+}][\ce{Cu^2+}][\ce{I^-}]}}$$ We rewrite the above equation after simplification:$$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{[\ce{Cu^+}][\ce{I^-}]}}$$ $$E^0_2(\ce{Cu^{2+}/CuI})= E_1^0(\ce{Cu^{2+}/Cu^+})+\frac{RT}{F}\ln{\frac{1}{K_{sp}(\ce{CuI})}}$$

$$E^0_2(\ce{Cu^{2+}/CuI})= 0.17+\frac{8.314 \times 298}{96500}\ln{\frac{1}{10^{-12}}}=0.88 \ce{V}$$

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