Here is a description of an analytical procedure for superconductors containing unknown quantities of Cu(I),Cu(II), Cu(III), and peroxide $\left(O_2^{2-}\right)$ : 33The possible trivalent copper and/or peroxide type oxygen are reduced by Cu(I) when dissolving the sample (ca .50 mg) in deoxygenated HCl solution ( 1 M) containing a known excess of monovalent copper ions (ca.25mgCuCI) . On the other hand, if the sample itself contained monovalent copper, the amount of Cu(I) in the solution would increase upon dissolving the sample. The excess Cu(I) was then determined by coulometric back titration... in an argon atmosphere." The abbreviation "ca." means "approximately." Coulometry is an electrochemical method in which the electrons liberated in the reaction$\mathit{C}{\mathit{u}}^{\mathbf{+}}\mathbf{\to }\mathit{C}{\mathit{u}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathit{e}$are measured from the charge flowing through an electrode. Explain with your own words and equations how this analysis works.

• The mole concept is a simple way to express the amount of a substance. Any measurement is divided into two parts: the numerical magnitude and the units in which the magnitude is expressed. For example, if the mass of a ball is 2 kilograms, the magnitude is '2' and the unit is 'kilogram.'
• When dealing with particles at the atomic (or molecular) level, even one gram of a pure element is known to contain a large number of atoms. This is a common application of the mole concept. It mainly concentrates on the component known as a mole,' which is a count of a very large number of particles.

To calculate moles of $C{o}^{3+}$that are contained in 25.00mg of material we need to calculate moles of $F{e}^{2+}$that reacted with$C{O}^{3+}$ .

To do that, we need to subtract initial moles of $F{e}^{2+}$and subtract them with excess (unreacted) moles of$F{e}^{2+}$ .

Initial moles of $F{e}^{2+}$are:

$$\begin{gathered} n\left(initialF{e}^{2+}\right)=c\left(F{e}^{2+}.V\left(F{e}^{2+}\right)\right) \\ n\left(initialF{e}^{2+}\right)=0.1000mol/L.5.000\times {10}^{-3}K \\ n\left(initialF{e}^{2+}\right)=5\times {10}^{-4}mol \end{gathered}$$

To equal the number of electrons on the left and right sides we need to multiply the second equation by 3 and then add reactants and products of each reaction together:

$$\begin{gathered} Reduction:C{r}_2{O}_7^{2-}+fre+14H^{+}\rightleftharpoons 2C{r}^{3+}+7H_{2}O \\ oxidation:6F{e}^{2+}\rightleftharpoons 6F{e}^{3+}+6e \end{gathered}$$

Net reaction is:

$Cr{}_2{O}_7^{2-}+14H^{+}+6f{e}^{2+}\rightleftharpoons 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

The number of moles that reacted with $C{O}^{3+}$are then:

$$\begin{gathered} n\left(F{e}^{2+}\right)=n\left(initialF{e}^{2+}\right)-n\left(excessF{e}^{2+}\right) \\ n\left(F{e}^{2+}\right)=5\times {10}^{-4}mol-3.085\times {10}^{-4}mol \\ n\left(F{e}^{2+}\right)=1.915\times {10}^{-4}mol \end{gathered}$$

By looking at the reaction of $C{o}^{3+}$and $F{e}^{2+}$the ratio they react in is 1:1 so their number of moles are equal. The number of mmoles of$C{O}^3$ in 25.00 mg sample is:

$n\left(C{o}^{3+}\right)=n\left(F{e}^{2+}\right)=1.915\times {10}^{-4}mol=0.1915mmol$

By subtracting total moles of Co with moles of $C{o}^{3+}$we can calculate moles of $C{o}^{2+}$:

$$\begin{gathered} n\left(C{o}^{2+}\right)=n\left(Co\right)-n\left(C{o}^{3+}\right) \\ n\left(C{o}^{2+}\right)=2.393\times {10}^{-4}mol-1.915\times {10}^{-4}mol=4.78\times {10}^{-5}mol \end{gathered}$$

Finally, we can calculate the average oxidation state by using this equation:

average oxidation state $=\frac{n\left(C{o}^{2+}\right).2+n\left(C{o}^{3+}\right).3}{n\left(totalCo\right)}$

average oxidation state $=\frac{4.78\times {10}^{-5}.2+1.915\times {10}^{-4}.3}{2.393\times {10}^{-4}}$

average oxidation state = 2.80

In the formula $L{i}_{1+y}Co{O}_2$, cobalt has an oxidation state 2.8 and oxygen - 4 so we can calculate y by writing this equation:

total oxidation state = Li + Co +2o

$$\begin{gathered} 0=Li+2.8-4 \\ Li=4-2.8 \\ Li=1.2 \end{gathered}$$

Since the oxidation state of lithium is 1.2 , the value of y is:

$$\begin{gathered} 1.2=1+y \\ y=1.2-1 \\ y=0.2 \end{gathered}$$

The theoretical weight percent is calculated by dividing the molar mass of the element with the total molar mass of the compound:

$$\begin{gathered} W\left(Li,L{i}_{1.2}Co{O}_2\right)=\frac{1.2M\left(Li\right)}{M\left(L{I}_{1.2}Co{O}_2\right)}.100\% \\ W\left(Li,L{i}_{1.2}Co{O}_2\right)=\frac{1.2.6.94g/moT}{\left(1.2.6.94+1.58.933+2.15.99\right)}.100\% \\ W\left(L{i}^2,L{i}_{1.2}Co{O}_2\right)=8.39\% \end{gathered}$$

Theoretical quotient wt\% Li/ wt\% Co is:

$$\begin{gathered} \frac{w\left(Li\right)}{w\left(Co\right)}=\frac{0.0839}{0.5938} \\ \frac{w\left(Li\right)}{w\left(Co\right)}=0.1429 \end{gathered}$$

The theoretical value is a little bit higher than the observed value of quotient.

Hence the working of this analysis is explained clearly.