91影视

• The degree of oxidation (i.e. the loss of electrons) of an atom inside a chemical compound is characterised by the oxidation number.
• The oxidation state and oxidation number are slightly different. The electronegativity of an atom in a bond should be included in the oxidation state.
• Electronegativity, on the other hand, is not taken into account when describing oxidation number.

(a)

Quantity of ${\mathrm{Co}}^{3+}$the presence of 25.00 mg It is necessary to explain some of the material.

To determine: the amount of${\mathrm{Co}}^{3+}$the presence of 25.00mg a representative sample

Initial ${\mathrm{Fe}}^{2+}$in $5.000\mathrm{mL}=(5.000\mathrm{mL})(0.100\mathrm{M})=0.500\mathrm{mmol}.$

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$unreacted titration requirements ${\mathrm{Fe}}^{2+}$

$=(3.228\mathrm{mL})(0.01593{\mathrm{MK}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7)=0.05142\mathrm{mmol}$

Immol, on the other hand,${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$interacts with six mols of hydrogen peroxide${\mathrm{Fe}}^{2+}$. This is a strong reaction,

${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+6{\mathrm{Fe}}^{2+}+14{\mathrm{H}}^{+}鈫�2{\mathrm{Cr}}^{3+}+6{\mathrm{Fe}}^{3+}+2{\mathrm{k}}^{+}+14{\mathrm{H}}_2\mathrm{O}$

So,${\mathrm{Fe}}^{2+}$ following the reaction with${\mathrm{Li}}_{1+\mathrm{y}}{\mathrm{CoO}}_2$

$$\begin{gathered} =\left(0.05142\mathrm{mmol}\right)\left(6{\mathrm{mmolFe}}^{2+}/{\mathrm{mmolK}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\right) \\ =0.3085{\mathrm{mmolCo}}^{3+}\mathrm{consumed} \\ {\mathrm{Fe}}^{2+}=\left(0.5000-0.3085\right) \\ =0.1915\mathrm{mmol} \end{gathered}$$

Imol Co $\mathrm{consumed}1{\mathrm{molFe}}^{2+}$, thereforelocalid="1667556359366" ${\mathrm{Co}}^{3+}$=0.1915mmol in a 25.00mg solid sample

(b)

Cobalt's average oxidation state must be established.

To find out what the average oxidation state of Co Cobalt is in 50.00mg solid = (0.564g Co/gsolid) (25.00 g solid) = 14.10mg

Cobalt in 25.00mg solid $=(14.10\mathrm{mg})(58.933\mathrm{g}/\mathrm{mol})=0.2393\mathrm{mmol}$

We know from component (a) that ${\mathrm{Co}}^{3+}=0.1915\mathrm{mmol}$, therefore

${\mathrm{Co}}^{2+}=0.2393-0.1915=0.0478\mathrm{mmol}$

Oxidation state of$\mathrm{Co}=\frac{(0.4987\mathrm{mmal})(2+)(0.1915\mathrm{mmol})(3+)}{0.2392\mathrm{mmol}}$

=2.80

Cobalt's average oxidation state is found to be 2.80.

(c)

In the formula, the value of y is${\mathrm{Li}}_{1+\mathrm{y}}{\mathrm{CoO}}_2$ It must be discovered.

To determine: y's value

Lithium should provide a charge of 1.20 if Cobalt has an average oxidation state of +2.80 and Oxygen has an oxidation state of -2, i.e. 4-2.80=1.20. As a result, the formula ${\mathrm{Li}}_{1+0.20}{\mathrm{CoO}}_2$and y has a value of 0.20.

(d)

Theoretical quotient weight percentage of L I in weight percentage of Co must be determined, and the observed quotient must be interpreted in terms of Cobalt's average oxidation state.

The mass percentage or weight percentage of a compound can be computed by multiplying the calculated mass of the substance by 100 to get the total mass of the sample.

The formula for calculating mass percent is Mass percent (inpercentage) $=\frac{(\mathrm{Culculated}\mathrm{mass}(\mathrm{ingrams})}{\left(\mathrm{totulmass}\right(\mathrm{ingrams})}脳100$

To determine: L i's weight percentage in Co's weight percentage Theoretical metal weight percentages in ${\mathrm{Li}}_1.20{\mathrm{CoO}}_2$is

Formula mass

$$\begin{gathered} =1.20(6.941)+1(58.933)+2(15.9994) \\ =99.261 \end{gathered}$$

Weight$\%ofLi=100脳1.20(6.941)/99.261=8.39\%$

Weightlocalid="1667556247196" $\%\mathrm{ofCo}=100脳58.933/99.261=59.37\%$

localid="1667556259358" $\frac{\mathrm{Weight}\%\mathrm{of}\mathrm{Li}}{\mathrm{Weight}\%\mathrm{ofCo}}=\frac{8.39\%}{59.37\%}=0.1413$

To explain: the measured fraction is consistent with Cobalt's average oxidation state. The observed ratio of Lithium weight % to Cobalt weight percentage is$0.1388卤0.0006.$The quotient calculated from the oxidation number is not exactly equal to this.