What are the oxidation numbers of the underlined elements in each of the following and how do you rationalize your results?

**(a) \( K\underline{I}_{3}\)**

**(b) \( H_{2}\underline{S}_{4}O_{6}\)**

**(c) \( \underline{Fe}_{3}O_{4}\)**

**(d) \( \underline{C}H_{3}\underline{C}H_{2}OH\)**

**(e) \( \underline{C}H_{3}\underline{C}OOH\)**

Asked by Abhisek | 1 year ago | 80

**(a)**** **\( K\underline{I}_{3}\)

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

1(+1) + 3(x) = 0

1 + 3x = 0

x = \( -\dfrac{1}{3}\)

Oxidation no. cannot be fractional. Hence, consider the structure of \( KI_{3}\).

In \( KI_{3}\) molecule, an iodine atom forms coordinate covalent bond with an iodine molecule.

Therefore, in \( KI_{3}\) molecule, the oxidation no. of I atoms forming the molecule \( I_{2}\) is 0, while the oxidation no. of I atom which is forming coordinate bond is -1.

Two of the four S atoms have an O.N. of +5, whereas the other two have an O.N. of 0

**(b)** \( H_{2}\underline{S}_{4}O_{6}\)

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 4(x) + 6(-2) = 0

2 + 4x -12 = 0

4x = 10

x = \( +2\dfrac{1}{2}\)

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule. The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

**(c)** \( \underline{Fe}_{3}O_{4}\)

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

3(x) + 4(-2) = 0

3x -8 = 0

x = \( \dfrac{8}{3}\)

Oxidation no. cannot be fractional.

One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.

\( FeO^{+2} Fe_2O_3\)

**(d)** \(\underline{C}H_{3}\underline{C}H_{2}OH\)

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 1(-2) = 0

2x + 4 -2 = 0

x = -2

Therefore, oxidation no. of C is -2.

**(e)** \( \underline{C}H_{3}\underline{C}OOH\)

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(x) + 4(+1) + 2(-2) = 0

2x + 4 -4 = 0

x = 0

Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in \( CH_{3}COOH\)

Answered by Abhisek | 1 year agoFluorine reacts with ice and results in the change:

\( H_{ 2 }O_{ (s) } \; + \; F_{ 2 \; (g) } \; \rightarrow \; HF_{ (g) } \; + \; HOF_{ (g) } \)

Justify that this reaction is a redox reaction

Depict the galvanic cell in which the reaction is:

\( Zn_{\left ( s \right )} \;+ \;2 \;Ag_{\left ( aq \right )}^{+} \;\rightarrow \;Zn^{2+}_{\left ( aq \right )} \;+ \;2 \;Ag_{\left ( s \right )} \)

Further show:

**(i)** which of the electrode is negatively charged?

**(ii)** the carriers of the current in the cell.

**(iii)** individual reaction at each electrode.

Given the standard electrode potentials,

\( \dfrac{ K^{+}}{K}\)= –2.93V

\( \dfrac{ Ag^{+}}{Ag}\) = 0.80V

\( \dfrac{Hg^{2+}}{Hg}\) = 0.79V

\(\dfrac{ Mg^{2+}}{Mg}\) = –2.37V

\(\dfrac{Cr^{3+}}{Cr}\) = –0.74V

Arrange these metals in their increasing order of reducing power.

Arrange the given metals in the order in which they displace each other from the solution of their salts.

Al, Fe, Cu, Zn, Mg

Predict the products of electrolysis in each of the following:

**(i) **An aqueous solution of \( AgNO_{3}\) with silver electrodes

**(ii)** An aqueous solution \( AgNO_{3}\) with platinum electrodes

**(iii)** A dilute solution of \( H_{2}SO_{4}\) with platinum electrodes

**(iv)** An aqueous solution of \( CuCl_{2}\)with platinum electrodes.