b) What do you mean by "Cathodic depolai
c) A $5.0 \mathrm{mM}$ solution of $\mathrm{Cd}(\mathrm{II})$ ion give the diffusion current of $40 \mu \mathrm{A}$ at constant temperature. Find the diffusion coefficient of Cd(II) ion. [Given : Rate of flow of mercury $=3.2 \mathrm{mg} \cdot \mathrm{s}^{-1}$, Drop time $=2.2 \mathrm{~s}$ )
$$
[3+(3+2)+2]
$$

Question

b) What do you mean by "Cathodic depolai
c) A $5.0 \mathrm{mM}$ solution of $\mathrm{Cd}(\mathrm{II})$ ion give the diffusion current of $40 \mu \mathrm{A}$ at constant temperature. Find the diffusion coefficient of Cd(II) ion. [Given : Rate of flow of mercury $=3.2 \mathrm{mg} \cdot \mathrm{s}^{-1}$, Drop time $=2.2 \mathrm{~s}$ )
$$
[3+(3+2)+2]
$$

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Identify the relevant equation**: The Ilkovic equation relates the diffusion current ($i_d$) to the diffusion coefficient ($D$) of the ion in a dropping mercury electrode (DME) system. The equation is given by: $i_d = 607nDC^{2/3}m^{1/6}t^{1/2}$, where $n$ is the number of electrons involved in the electrode reaction, $D$ is the diffusion coefficient in $\mathrm{cm}^2 \mathrm{s}^{-1}$, $C$ is the concentration of the depolarizer in $\mathrm{mol} \mathrm{cm}^{-3}$, $m$ is the rate of flow of mercury in $\mathrm{mg} \mathrm{s}^{-1}$, and $t$ is the drop time in seconds. ‖ ‖ 2 ⋮ **Convert given units and plug in the values**: Convert the concentration from $\mathrm{mM}$ to $\mathrm{mol} \mathrm{cm}^{-3}$, the current from $\mu \mathrm{A}$ to $\mathrm{A}$, and the flow rate from $\mathrm{mg} \mathrm{s}^{-1}$ to $\mathrm{g} \mathrm{s}^{-1}$. Then, plug these values into the Ilkovic equation to solve for $D$. Given values are: $C = 5.0 \mathrm{mM} = 5.0 imes 10^{-3} \mathrm{mol} \mathrm{L}^{-1} = 5.0 imes 10^{-6} \mathrm{mol} \mathrm{cm}^{-3}$, $i_d = 40 \mu \mathrm{A} = 40 imes 10^{-6} \mathrm{A}$, $m = 3.2 \mathrm{mg} \mathrm{s}^{-1} = 3.2 imes 10^{-3} \mathrm{g} \mathrm{s}^{-1}$, and $t = 2.2 \mathrm{s}$. For $\mathrm{Cd}(\mathrm{II})$ ions, $n = 2$. ‖ ‖ 3 ⋮ **Solve for the diffusion coefficient $D$**: Rearrange the Ilkovic equation to solve for $D$ and substitute the known values: $D = \frac{i_d}{607 \cdot n \cdot C^{2/3} \cdot m^{1/6} \cdot t^{1/2}}$. ‖ ‖ 4 ⋮ **Calculate the diffusion coefficient**: Using the rearranged equation and the converted values, calculate $D$: $D = \frac{40 imes 10^{-6} \mathrm{A}}{607 \cdot 2 \cdot (5.0 imes 10^{-6} \mathrm{mol} \mathrm{cm}^{-3})^{2/3} \cdot (3.2 imes 10^{-3} \mathrm{g} \mathrm{s}^{-1})^{1/6} \cdot (2.2 \mathrm{s})^{1/2}}$. ‖ ∻Answer∻ ⚹ $D = 7.96 imes 10^{-6} \mathrm{cm}^2 \mathrm{s}^{-1}$ ⚹ ∻Key Concept∻ ⚹ The Ilkovic equation is used to calculate the diffusion coefficient of an ion in a dropping mercury electrode system. ⚹ ∻Explanation∻ ⚹ The diffusion coefficient is found by rearranging the Ilkovic equation and substituting the known values of the diffusion current, concentration, flow rate of mercury, and drop time. ⚹◊What is the relationship between diffusion current and the diffusion coefficient in electrochemistry?⍭ Generate me a similar question◊

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