Question

A solution containing $c{d}^2+$was analyzed voltammetrically using the standard addition method. Twenty-five milliliters of the deaerated solution, which was $1M$in $HN{O}_3$, produced a net limiting current of $1.41\mu \mathrm{\{}A$} at a rotating mercury film working electrode at a potential of $-0.8V$(versus $SCE$). Following addition of $5.00ML$of a $2.50\times {10}^{-3}\mathrm{M}$standard ${\mathrm{Cd}}^{2+}$solution, the resulting solution produced a current of $4.93\mu \mathrm{\{}A$}. Calculate the concentration of ${\mathrm{Cd}}^{2+}$the sample.

Short answer

The concentration of$c{d}^2+$is,$1.56\times {10}^{-4}$

Step-by-step solution

Current system.

The current of a system at any point in electrolysis is determined by the rate of transport of diffusion to the electrode surface. The relation between limiting current and concentration of an analyte can be written as follows:

$i_1=k_{\mathrm{A}}{c}_{\mathrm{A}}$

Here$,i_{\{}l$} is limiting current, $k$is the constant for reactant, and $c_A$is the reactant concentration.

Containing.

The limiting current for the$25ml$solution containing ${\mathrm{Cd}}^{2+}$can be expressed as follows:

role="math" localid="1650436538859" $i_1=k{\left(c_{{\mathrm{Cd}}^{2*}}\right)}_{}\dots \dots \left(1\right)$

On the addition of $5.00ml$of $2.5\times {10}^{-3}{\mathrm{MCd}}^{2+}$to the initial solution, the concentration of ${\mathrm{Cd}}^{2+}$changes as follows:

$\left(c_{{\left.{\mathrm{Cd}}^{2*}\right)}_2}=\frac{25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}}{25\mathrm{ml}+5\mathrm{ml}}\right.$

$=\frac{25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}}{30\mathrm{ml}}$

The limiting current for the solution after the addition of $5.00\mathrm{ml}$of $2.5\times {10}^{-3}{\mathrm{MCd}}^{2+}$is as follows:

$i_2=k{\left(c_{{\mathrm{Cd}}^{2*}}\right)}_2$

Substitute the value of concentration in the above expression as follows:

$i_2=k\frac{\left(25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}\right)}{30\mathrm{ml}}$

Substitute the value of $k$by the equation $\left(1\right)$is as follows:

$i_2=\frac{i_1}{{\left(c_{{\mathrm{Cd}}^{3+}}\right)}_1}\times \frac{\left(25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}\right)}{30\mathrm{ml}}$

Substitute $i_1$as $1.41\mu \mathrm{A},i_2$as $4.93\mu \mathrm{A}$and ${\left(c_{{\mathrm{Cd}}^{2*}}\right)}_1$as $\left[{\mathrm{Cd}}^{2+}\right]$in the above expression.

$4.93\mu \mathrm{A}=\frac{1.41\mu \mathrm{A}}{\left[{\mathrm{Cd}}^{2+}\right]}\times \frac{\left(25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}\right)}{30\mathrm{ml}}$

Expression.

Solve the above expression as follows:

$(4.93\mu \mathrm{A})\left(30\mathrm{ml}\right)\left[{\mathrm{Cd}}^{2+}\right]=(1.41\mu \mathrm{A})\left(25\mathrm{ml}\times \left[{\mathrm{Cd}}^{2+}\right]+5\mathrm{ml}\times 2.50\times {10}^{-3}\mathrm{M}\right)$or $(147.9-35.25)\left[{\mathrm{Cd}}^{2+}\right]=17.625\mathrm{M}$or $\left[{\mathrm{Cd}}^{2+}\right]=\frac{17.625\mathrm{M}}{112.65}$

$=0.156\times {10}^{-3}\mathrm{M}$

$=1.56\times {10}^4\mathrm{M}$