Question

Hydrogen sulphide $\left( {{H_2}S} \right)$ is a strong reducing agent. Which of the following reactions shows its reducing action

(A) $Cd{\left( {N{O_3}} \right)_2} + \;{H_2}S \to CdS + 2HNO{_3}$

(B) $CuS{O_4} + {H_2}S \to CuS + {H_2}S{O_4}$

(C) $2FeC{l_3} + {H_2}S \to 2FeC{l_2} + 2HCl + S$

(D) $Pb{\left( {N{O_3}} \right)_2} + {H_2}S \to PbS + 2C{H_3}COOH$

Answer 1

We need to know what a reducing agent is. A reducing agent is an element or a compound that donates electrons to an electron acceptor(oxidising agent). In a redox reaction where oxidation and reduction take place simultaneously, a reducing agent gets oxidised when it donates electrons to an oxidising agent which in turn gets reduced.

Complete step by step answer:

As given in the question, ${H_2}S$ is a strong reducing agent. To identify its reducing action in a reaction, its oxidation state must increase. We now calculate the oxidation state of ${H_2}S$ before and after the reaction using the rules above for each of the given reactions.

(A) $Cd{\left( {N{O_3}} \right)_2} + {H_2}S \to CdS + 2HN{O_3}$

In the reactant: In the compound ${H_2}S$ , the two hydrogen atoms have an oxidation state of $+ 1$ each.Hence the oxidation state of ${H_2}$ is $+ 2$ and that of $S$ is $- 2$ (The sum of their oxidation states must be equal to zero as it is a neutral molecule).

In the product: In The molecule $CdS$ , the oxidation state of $Cd$ is $+ 2$ (group 12 element) and that of $S$ is $- 2$ . In the molecule $2HN{O_3}$ , 2 hydrogens have an oxidation state of $+ 2$ .

Since the oxidation state of Hydrogen and Sulphur did not change before and after the reaction, hence it does not show its reducing action. Therefore, option A is incorrect.

(B) $CuS{O_4} + {H_2}S \to CuS + {H_2}S{O_4}$

In the reactant: In the compound ${H_2}S$ , the two hydrogen atoms have an oxidation state of $+ 1$ each.Hence the oxidation state of ${H_2}$ is $+ 2$ and that of $S$ is $- 2$ (The sum of their oxidation states must be equal to zero as it is a neutral molecule).

In the product: In The molecule CuS, Cu has an oxidation state of $+ 2$ and thus $S$ has $- 2$ . In the molecule ${H_2}S{O_4}$ ,2 hydrogens have an oxidation state of $+ 2$ .

Since the oxidation state of Hydrogen and Sulphur did not change before and after the reaction, hence it does not show its reducing action. Therefore, option B is incorrect.

(C) $2FeC{l_3} + {H_2}S \to 2FeC{l_2} + 2HCl + S$

In the reactant: In the compound ${H_2}S$ , the two hydrogen atoms have an oxidation state of $+ 1$ each.Hence the oxidation state of ${H_2}$ is $+ 2$ and that of $S$ is $- 2$ (The sum of their oxidation states must be equal to zero as it is a neutral molecule).

In the product: In the molecule $2HCl$ , 2 hydrogens have an oxidation state of $+ 2$ . The oxidation state of $S$ is $0$ since it is a free element.

Since the oxidation state of sulphur increases from $- 2$ to $0$ before and after the reaction, hence it can be said that ${H_2}S$ shows its reducing action in this reaction. Therefore, option C is correct.

(D) $Pb{\left( {N{O_3}} \right)_2} + {H_2}S \to PbS + 2C{H_3}COOH$

In the reactant: In the compound ${H_2}S$ ,The two hydrogen atoms have an oxidation state of $+ 1$ each.Hence the oxidation state of ${H_2}$ is $+ 2$ and that of $S$ is $- 2$ (The sum of their oxidation states must be equal to zero as it is a neutral molecule).

In the product: In the molecule $PbS$ ,the oxidation state of $S$ is $- 2$ . In the molecule $C{H_3}COOH$ ,each Hydrogen atom has an oxidation state of $+ 1$ .

Since the oxidation state of Hydrogen and Sulphur did not change before and after the reaction, hence it does not show its reducing action. Therefore, option D is incorrect.

Therefore, the option (C) is correct.

Additional information:

We must know that the reducing agents are the elements or compounds that are oxidized (oxidation state increases) and oxidizing agents are the elements or compounds that are reduced (oxidation state decreases).

To identify the reducing action of a reducing agent in a reaction, we must calculate the oxidation state or oxidation number of the element or the compound. The oxidation state must increase in order to exhibit reducing action. The oxidation state sometimes referred to as oxidation number, describes the degree of oxidation(loss of electrons) of an atom in a chemical compound.

Some general rules are to be followed when calculating the oxidation number of any element. They are discussed as follows:

(i) In a neutral compound, all oxidation numbers must add up to zero. A neutral compound does not have a plus or minus charge.

(ii)In an ion, all the oxidation numbers must add up to the charge of the ion.

(iii) Free elements have an oxidation number of zero (e.g. $Na$ , $Fe$ , $H_2$ , $O_2$ , and $S_8$ ).

(iv) Another straightforward rule is Fluorine always has its oxidation number as $- 1$ it is the most electronegative element and has one electron less to complete its octet.

(v)Elements in the Group 1 of the periodic table have an oxidation number of $+ 1$ and those of group 2 elements have an oxidation number of $+ 2$.

(vi) Hydrogen, when bonded to nonmetals, has an oxidation number of +1 and hydrogen when bonded to metals (or Boron) have an oxidation number of $- 1$.

(vii) Oxygen usually has an oxidation state of $- 2$ with two big exceptions,i.e, when bonded to Fluorine and in peroxides.

(viii)When none of the above rules seems applicable, we can use the group rules. In binary compounds with metals,

Group 17 elements have oxidation state $- 1$.

Group 16 elements have oxidation state $- 2$.

Group 15 elements have oxidation state $- 3$.

Note: We need to keep in mind that for calculating the pH value of a solution of a weak base it is similar to that of the weak acid. However, the variable x in the solution will represent the concentration of the hydroxide ion. For finding the pH we have to take the negative logarithm for getting the pOH value followed by subtracting from 14 to get the pH i.e. pH = 14-1 =13.