Question

How many of the following species can be reduced by $SO_2$ in aqueous or acidic medium?

(i) HOCl (ii) $HgCl_2$ (iii) $H_2O_2$ (iv) $KMnO_4$ (v) $SnCl_4$ (vi) $Fe_2(SO_4)_3$ (vii) $KIO_3$ (viii) $H_2S$ (ix) $Cl_2$ (x) $AgNO_3$

Answer 1

8

Answer 2

To determine which species can be reduced by $SO_2$ in aqueous or acidic medium, we need to consider the redox properties of $SO_2$.

When $SO_2$ acts as a reducing agent, it gets oxidized. In aqueous or acidic medium, $SO_2$ is typically oxidized to sulfate ($SO_4^{2-}$). The half-reaction for this oxidation is:

$SO_2(g) + 2H_2O(l) \rightarrow SO_4^{2-}(aq) + 4H^+(aq) + 2e^-$

The standard reduction potential for the reverse reaction, $SO_4^{2-}(aq) + 4H^+(aq) + 2e^- \rightarrow SO_2(g) + 2H_2O(l)$, is $E^\circ = 0.17 V$.

For $SO_2$ to reduce another species (let's call it $Ox$), the overall reaction $Ox + SO_2 \rightarrow Red + SO_4^{2-}$ must be spontaneous. This means the standard cell potential ($E^\circ_{cell}$) for this reaction must be positive.

$E^\circ_{cell} = E^\circ_{reduction}(Ox/Red) - E^\circ_{reduction}(SO_4^{2-}/SO_2)$

For $E^\circ_{cell} > 0$, we must have $E^\circ_{reduction}(Ox/Red) > E^\circ_{reduction}(SO_4^{2-}/SO_2)$, which means $E^\circ_{reduction}(Ox/Red) > 0.17 V$.

Additionally, a species must be in an oxidation state higher than its lowest possible oxidation state to be reduced.

Let's examine each species:

(i) HOCl (Hypochlorous acid): Chlorine is in +1 oxidation state. It can be reduced to $Cl^-$ (-1). $E^\circ(HOCl/Cl^-) = 1.49 V$. Since $1.49 V > 0.17 V$, HOCl can be reduced.

(ii) $HgCl_2$ (Mercury(II) chloride): Mercury is in +2 oxidation state. It can be reduced to $Hg$ (0) or $Hg_2Cl_2$ (Hg in +1). $E^\circ(HgCl_2/Hg) = 0.63 V$. Since $0.63 V > 0.17 V$, $HgCl_2$ can be reduced.

(iii) $H_2O_2$ (Hydrogen peroxide): Oxygen is in -1 oxidation state. It can be reduced to $H_2O$ (oxygen in -2). $E^\circ(H_2O_2/H_2O) = 1.77 V$. Since $1.77 V > 0.17 V$, $H_2O_2$ can be reduced.

(iv) $KMnO_4$ (Potassium permanganate): Manganese is in +7 oxidation state. It can be reduced to $Mn^{2+}$ (+2) in acidic medium. $E^\circ(MnO_4^-/Mn^{2+}) = 1.51 V$. Since $1.51 V > 0.17 V$, $KMnO_4$ can be reduced.

(v) $SnCl_4$ (Tin(IV) chloride): Tin is in +4 oxidation state. It can be reduced to $Sn^{2+}$ (+2). $E^\circ(Sn^{4+}/Sn^{2+}) = 0.15 V$. Since $0.15 V < 0.17 V$, $SnCl_4$ cannot be reduced by $SO_2$.

(vi) $Fe_2(SO_4)_3$ (Iron(III) sulfate): Iron is in +3 oxidation state. It can be reduced to $Fe^{2+}$ (+2). $E^\circ(Fe^{3+}/Fe^{2+}) = 0.77 V$. Since $0.77 V > 0.17 V$, $Fe_2(SO_4)_3$ can be reduced.

(vii) $KIO_3$ (Potassium iodate): Iodine is in +5 oxidation state. It can be reduced to $I_2$ (0) or $I^-$ (-1). $E^\circ(IO_3^-/I_2) = 1.20 V$ (or $E^\circ(IO_3^-/I^-) = 1.08 V$). Since both are $> 0.17 V$, $KIO_3$ can be reduced.

(viii) $H_2S$ (Hydrogen sulfide): Sulfur is in -2 oxidation state. This is the lowest possible oxidation state for sulfur, so $H_2S$ cannot be reduced further.

(ix) $Cl_2$ (Chlorine gas): Chlorine is in 0 oxidation state. It can be reduced to $Cl^-$ (-1). $E^\circ(Cl_2/Cl^-) = 1.36 V$. Since $1.36 V > 0.17 V$, $Cl_2$ can be reduced.

(x) $AgNO_3$ (Silver nitrate): Silver is in +1 oxidation state. It can be reduced to $Ag$ (0). $E^\circ(Ag^+/Ag) = 0.80 V$. Since $0.80 V > 0.17 V$, $AgNO_3$ can be reduced.

Counting the species that can be reduced: (i), (ii), (iii), (iv), (vi), (vii), (ix), (x). Total number of species is 8.

Explanation of the solution:

$SO_2$ acts as a reducing agent by getting oxidized to $SO_4^{2-}$ (sulfur changes from +4 to +6 oxidation state). The standard reduction potential for $SO_4^{2-}/SO_2$ is $0.17 V$. For $SO_2$ to reduce another species, that species must have a standard reduction potential greater than $0.17 V$ and must be in an oxidation state that allows for reduction.

1. HOCl: Cl(+1) can be reduced. $E^\circ = 1.49 V > 0.17 V$. (Yes)
2. $HgCl_2$: Hg(+2) can be reduced. $E^\circ = 0.63 V > 0.17 V$. (Yes)
3. $H_2O_2$: O(-1) can be reduced. $E^\circ = 1.77 V > 0.17 V$. (Yes)
4. $KMnO_4$: Mn(+7) can be reduced. $E^\circ = 1.51 V > 0.17 V$. (Yes)
5. $SnCl_4$: Sn(+4) can be reduced. $E^\circ = 0.15 V < 0.17 V$. (No)
6. $Fe_2(SO_4)_3$: Fe(+3) can be reduced. $E^\circ = 0.77 V > 0.17 V$. (Yes)
7. $KIO_3$: I(+5) can be reduced. $E^\circ = 1.08 V > 0.17 V$. (Yes)
8. $H_2S$: S(-2) is in its lowest oxidation state, cannot be reduced. (No)
9. $Cl_2$: Cl(0) can be reduced. $E^\circ = 1.36 V > 0.17 V$. (Yes)
10. $AgNO_3$: Ag(+1) can be reduced. $E^\circ = 0.80 V > 0.17 V$. (Yes)

Total species that can be reduced: 8.