Question

Suppose that the diffusion current in a polarogram for reduction of ${\mathrm{Cd}}^{2+}$at a mercury electrode is$14\mathrm{\mu A}$If the solution containsof 25mLof$0.50{\mathrm{mMCd}}^{2+}$what percentage of${\mathrm{Cd}}^{2+}$is reduced in the 3.4 min required to scan from$-0.6\mathbf{to}-1.2\mathrm{V}$?

Short answer

The percentage of cadmium ion which can be reduced in polarogram is 0.12%.

Step-by-step solution

Define Molarity.

At any point in an electric circuit the charge is the product of current I and time.

q = l . t

The number of moles of electronsreactreacts in a reaction with a given timeis

Reacted moles$=\frac{\mathrm{l}.\mathrm{t}}{\mathrm{nF}}$

 Determine the Molarity of Cadmium.

Given,

$$\begin{gathered} \mathrm{l}=14\times {10}^{-6}\mathrm{A} \\ \mathrm{t}=3.4\min \\ =204\sec \end{gathered}$$

Number of moles of electrons is given as

role="math" $=\frac{(14\times {10}^{-6}\mathrm{A})\left(204\sec \right)}{96485\mathrm{C}/\mathrm{mol}}=2.9\times {10}^{-8}\mathrm{mol}$

The reduced cadmium ion is calculated as $=\frac{2.9\times {10}^{-8}\mathrm{mol}}{2}=1.48\times {10}^{-8}\mathrm{mol}$

The 25mL of cadmium ions with has $=\frac{25\times 0.5\times {10}^{-3}}{1000}=1.25\times {10}^{-5}\mathrm{moles}$

Reduced \% of cadmium ions$=\frac{1.48\times {10}^{-8}\mathrm{moles}}{1.25\times {10}^{-5}\mathrm{moles}}\times 100$