The diffusion coefficient of $\mathrm{Ti}$ is $2.0 imes 10^{-5} \mathrm{~cm}^{2} \mathrm{~s}^{-1}$ and that for $\mathrm{Cd}^{2+}$ is $0.72 imes 10^{-5} \mathrm{~cm}^{2} \mathrm{~s}^{-1}$. If a I $\mathrm{mM}$ solution of $\mathrm{Cd}^{2+}$ gives a diffusion current of $8.15 \mu \mathrm{A}$, what will be the diffusion current for a $1 \mathrm{mM}$ solution of $T i^{+}$using the same polarographic cell?

Question

The diffusion coefficient of $\mathrm{Ti}$ is $2.0 	imes 10^{-5} \mathrm{~cm}^{2} \mathrm{~s}^{-1}$ and that for $\mathrm{Cd}^{2+}$ is $0.72 	imes 10^{-5} \mathrm{~cm}^{2} \mathrm{~s}^{-1}$. If a I $\mathrm{mM}$ solution of $\mathrm{Cd}^{2+}$ gives a diffusion current of $8.15 \mu \mathrm{A}$, what will be the diffusion current for a $1 \mathrm{mM}$ solution of $T i^{+}$using the same polarographic cell?

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Apply the Ilkovic equation**: The diffusion current ($I_d$) for a reversible, Nernstian, planar diffusion-controlled electrochemical process is given by the Ilkovic equation: $I_d = nFAD^{2/3}Cv^{1/6}t^{1/2}$, where $n$ is the number of electrons transferred, $F$ is the Faraday constant, $A$ is the electrode area, $D$ is the diffusion coefficient, $C$ is the concentration, $v$ is the kinematic viscosity, and $t$ is the drop time of the mercury electrode. ‖ ‖ 2 ⋮ **Calculate the diffusion current for $\mathrm{Ti}^{+}$**: Given that the diffusion current for $\mathrm{Cd}^{2+}$ is $8.15 \mu \mathrm{A}$, we can use the ratio of the diffusion coefficients to find the diffusion current for $\mathrm{Ti}^{+}$. Since all other factors are constant and the concentration is the same ($1 \mathrm{mM}$), the diffusion current is directly proportional to the diffusion coefficient to the power of $2/3$. Thus, $I_{d,\mathrm{Ti}} = I_{d,\mathrm{Cd}} \left( \frac{D_{\mathrm{Ti}}}{D_{\mathrm{Cd}}} ight)^{2/3}$. ‖ ‖ 3 ⋮ **Perform the calculation**: $I_{d,\mathrm{Ti}} = 8.15 \mu \mathrm{A} \left( \frac{2.0 imes 10^{-5}}{0.72 imes 10^{-5}} ight)^{2/3}$. ‖ ‖ 4 ⋮ **Solve for $I_{d,\mathrm{Ti}}$**: After calculating the above expression, we get the diffusion current for $\mathrm{Ti}^{+}$. ‖ ∻Answer∻ ⚹ $I_{d,\mathrm{Ti}} \approx 13.9 \mu \mathrm{A}$ ⚹ ∻Key Concept∻ ⚹ Diffusion current in polarography is proportional to the diffusion coefficient to the power of $2/3$ when all other variables are constant. ⚹ ∻Explanation∻ ⚹ The diffusion current for $\mathrm{Ti}^{+}$ is calculated using the Ilkovic equation and the known diffusion current for $\mathrm{Cd}^{2+}$, taking into account the different diffusion coefficients of $\mathrm{Ti}$ and $\mathrm{Cd}^{2+}$. ⚹◊What is the relationship between diffusion coefficient and diffusion current in polarography?⍭ Generate me a similar question◊

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