Mix Examples-Redox Reactions Questions in English

(C) Acidified $KMnO_4$ acts as a strong oxidizing agent.
$2KMnO_4 + 3H_2SO_4 \to K_2SO_4 + 2MnSO_4 + 3H_2O + 5[O]$
$Mohr's$ salt,which is $FeSO_4(NH_4)_2SO_4 \cdot 6H_2O$,contains $Fe^{2+}$ ions which can be oxidized to $Fe^{3+}$ ions.
Therefore,$Mohr's$ salt acts as a reducing agent and decolourises the purple solution of acidified $KMnO_4$.

(D) From the balanced chemical equation:
$MnO_4^- + 5Fe^{2+} + 8H^+ \to Mn^{2+} + 5Fe^{3+} + 4H_2O$
$1 \ \text{mole of } KMnO_4 \text{ reacts with } 5 \ \text{moles of } Fe^{2+}$.
Using the stoichiometry: $n_{Fe^{2+}} = 5 \times n_{KMnO_4}$
$M_1 V_1 = 0.1 \ M \times 10 \ mL = 1 \ \text{mmol of } KMnO_4$.
Therefore,moles of $Fe^{2+}$ required = $5 \times 1 \ \text{mmol} = 5 \ \text{mmol}$.
For $0.1 \ M$ $FeSO_4$ solution,$V = \frac{\text{moles}}{\text{Molarity}} = \frac{5 \ \text{mmol}}{0.1 \ M} = 50 \ mL$.
Thus,$10 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to $50 \ mL$ of $0.1 \ M$ $FeSO_4$.

(D) The balanced redox reaction in an acidic medium is:
$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$
According to the law of equivalence,the number of equivalents of $KMnO_4$ must equal the number of equivalents of $FeSO_4$.
Equivalents = $Molarity \times n-factor \times Volume(L)$.
For $FeSO_4$,the $n-factor$ is $1$ $(Fe^{2+} \rightarrow Fe^{3+} + e^-)$.
For $KMnO_4$,the $n-factor$ is $5$ $(MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O)$.
Equivalents of $FeSO_4 = \frac{1}{10} \times 1 \times 10 \ mL = 1 \ meq$.
Equivalents of $KMnO_4 = M \times 5 \times V = 1 \ meq$.
Checking option $(d)$: $0.02 \ M \times 5 \times 10 \ mL = 1 \ meq$.
Thus,the correct option is $(d)$.

(B) The reaction is: $\text{SO}_2 + 2\text{H}_2\text{S} \to 3\text{S} + 2\text{H}_2\text{O}$.
In this reaction,$\text{SO}_2$ acts as the oxidising agent where the oxidation state of $S$ changes from $+4$ to $0$.
The change in oxidation state is $4 - 0 = 4$.
The equivalent mass $(EW)$ of $\text{SO}_2$ is $\frac{\text{Molar Mass}}{n\text{-factor}} = \frac{64}{4} = 16$.
The value twice the equivalent mass is $16 \times 2 = 32$.

(A) The oxidation state of $N$ in $NO_3^-$ is $+5$.
To find the number of moles of electrons taken up,we calculate the change in oxidation state for each reduction product:
$1$. For $NO_3^- \rightarrow NH_3$: Oxidation state of $N$ changes from $+5$ to $-3$. Electrons taken $= 5 - (-3) = 8 \ mol$.
$2$. For $NO_3^- \rightarrow NH_2OH$: Oxidation state of $N$ changes from $+5$ to $-1$. Electrons taken $= 5 - (-1) = 6 \ mol$.
$3$. For $NO_3^- \rightarrow NO$: Oxidation state of $N$ changes from $+5$ to $+2$. Electrons taken $= 5 - 2 = 3 \ mol$.
$4$. For $NO_3^- \rightarrow NO_2$: Oxidation state of $N$ changes from $+5$ to $+4$. Electrons taken $= 5 - 4 = 1 \ mol$.
The maximum number of moles of electrons taken up is $8 \ mol$ (when reduced to $NH_3$).

(B) The given reaction is $4Fe + 3O_2 \to 4Fe^{3+} + 6O^{2-}$.
In this reaction,the oxidation state of $Fe$ increases from $0$ to $+3$,which means $Fe$ is oxidised.
Since $Fe$ is oxidised,it acts as a reducing agent.
Conversely,the oxidation state of $O$ decreases from $0$ to $-2$,meaning $O_2$ is reduced.
Statement $B$ is incorrect because metallic iron is oxidised to $Fe^{3+}$,not reduced.

(D) In the given reaction: $4Fe^{0} + 3O_2^{0} \to 4Fe^{3+} + 6O^{2-}$.
$Fe$ is oxidized from $0$ to $+3$ state,so it acts as a reducing agent.
$O_2$ is reduced from $0$ to $-2$ state,so it acts as an oxidizing agent.
Since both oxidation and reduction occur,it is a redox reaction.
Statement $D$ is incorrect because metallic iron is oxidized to $Fe^{3+}$,not reduced.

(D) In the given reaction: $2CrO_4^{2-} + 2H^{+} \to Cr_2O_7^{2-} + H_2O$,we observe the oxidation states of the elements involved.
For $Cr$ in $CrO_4^{2-}$,the oxidation state is $x + 4(-2) = -2$,so $x = +6$.
For $Cr$ in $Cr_2O_7^{2-}$,the oxidation state is $2x + 7(-2) = -2$,so $2x = +12$,$x = +6$.
Since there is no change in the oxidation state of any element,this is not a redox reaction.
Therefore,there is no oxidising agent or reducing agent present in this reaction.

(C) The reaction between $CuSO_4$ and $KI$ is as follows:
$2CuSO_4 + 4KI \rightarrow 2K_2SO_4 + 2CuI_2$
$2CuI_2 \rightarrow Cu_2I_2 + I_2$
In $CuSO_4$,the oxidation state of $Cu$ is $+2$.
In $Cu_2I_2$ (which is $CuI$),the oxidation state of $Cu$ is $+1$.
Therefore,the change in the oxidation number of $Cu$ is $|(+1) - (+2)| = 1$.

(C) In the reaction $BaO_2 + H_2SO_4 \to BaSO_4 + H_2O_2$,we analyze the oxidation states of the elements involved.
In $BaO_2$ (barium peroxide),$Ba$ is $+2$ and $O$ is $-1$.
In $H_2SO_4$,$H$ is $+1$,$S$ is $+6$,and $O$ is $-2$.
In $BaSO_4$,$Ba$ is $+2$,$S$ is $+6$,and $O$ is $-2$.
In $H_2O_2$ (hydrogen peroxide),$H$ is $+1$ and $O$ is $-1$.
Since the oxidation states of all elements $(Ba, O, H, S)$ remain unchanged on both sides of the reaction,this is not a redox reaction and there is no change in valency.

(A) In the given reaction,$I_2$ is reduced to $2I^-$.
The oxidation state of iodine changes from $0$ to $-1$.
Total change in oxidation state per molecule of $I_2$ = $|2 \times (-1) - 0| = 2$.
Equivalent weight = $\frac{\text{Molecular weight}}{\text{n-factor}}$.
Since the n-factor is $2$,the equivalent weight of $I_2$ is $\frac{\text{Molecular weight}}{2}$.

(D) In $H_2S$,the oxidation state of sulphur is $-2$,which is its minimum possible oxidation state,so it can only be oxidized (act as a reducing agent).
In $SO_2$,the oxidation state of sulphur is $+4$,which is intermediate between its minimum $(-2)$ and maximum $(+6)$ oxidation states,allowing it to act as both an oxidizing agent (by decreasing its oxidation state) and a reducing agent (by increasing its oxidation state).

(D) In the reaction $HCl + H_2O \to H_3O^{+} + Cl^{-}$,there is no change in the oxidation state of any of the atoms involved.
Since there is no oxidation or reduction occurring,it is not a redox reaction.

(C) The correct answer is $(C)$.
Acidified potassium permanganate $(KMnO_4)$ acts as a strong oxidizing agent.
Mohr salt,which is $(FeSO_4)(NH_4)_2SO_4 \cdot 6H_2O$,contains $Fe^{2+}$ ions.
These $Fe^{2+}$ ions are oxidized to $Fe^{3+}$ ions by the acidified $KMnO_4$ solution,causing the purple color of the permanganate solution to disappear (decolourise).

(C) $HNO_2$ acts as both an oxidising and a reducing agent.
It acts as a reducing agent when it reacts with strong oxidising agents like $KMnO_4$,$K_2Cr_2O_7$,and $H_2O_2$.
It acts as an oxidising agent when it reacts with strong reducing agents like $HI$ and $H_2SO_3$.
For example,it oxidises $Fe^{+2}$ to $Fe^{+3}$:
$Fe^{+2} + HNO_2 + H^{+} \rightarrow Fe^{+3} + NO + H_2O$
It reduces acidified $KMnO_4$:
$2KMnO_4 + 3H_2SO_4 + 5HNO_2 \rightarrow K_2SO_4 + 2MnSO_4 + 3H_2O + 5HNO_3$

(A) $SO_2$ acts as a reducing agent in the presence of strong oxidizing agents and as an oxidizing agent in the presence of strong reducing agents.
$1.$ $SO_2$ oxidizes $Mg$ to $MgO$: $SO_2 + 2Mg \to 2MgO + S$.
$2.$ $SO_2$ reduces $K_2Cr_2O_7$ (in acidic medium) to $Cr^{3+}$: $Cr_2O_7^{2-} + 3SO_2 + 2H^+ \to 2Cr^{3+} + 3SO_4^{2-} + H_2O$.
$3.$ $SO_2$ reduces $KMnO_4$ (in acidic medium) to $Mn^{2+}$: $2MnO_4^- + 5SO_2 + 2H_2O \to 2Mn^{2+} + 5SO_4^{2-} + 4H^+$.
Since $SO_2$ acts as an oxidizing agent only for $Mg$ and as a reducing agent for $K_2Cr_2O_7$ and $KMnO_4$,the correct answer is $A$.

(B) The reaction between sodium thiosulfate $(Na_2S_2O_3)$,chlorine $(Cl_2)$,and water $(H_2O)$ is a redox reaction where chlorine acts as an oxidizing agent.
The balanced chemical equation is:
$Na_2S_2O_3 + 4Cl_2 + 5H_2O \to Na_2SO_4 + 8HCl + S$
Thus,the products formed are sulfur $(S)$,hydrochloric acid $(HCl)$,and sodium sulfate $(Na_2SO_4)$.

(C) The iodine solution $(I_2)$ reacts with sodium thiosulphate $(Na_2S_2O_3)$ to form colourless sodium iodide $(NaI)$ and sodium tetrathionate $(Na_2S_4O_6)$.
The chemical equation is: $2Na_2S_2O_3 + I_2 \to 2NaI + Na_2S_4O_6$.
Since the product $NaI$ is colourless,the brown colour of the iodine solution disappears.

(B) The reaction between bromine water ($Br_2$ in $H_2O$) and sulfur dioxide $(SO_2)$ is a redox reaction where $SO_2$ acts as a reducing agent and $Br_2$ acts as an oxidizing agent.
The balanced chemical equation is:
$Br_2 + 2H_2O + SO_2 \rightarrow H_2SO_4 + 2HBr$
Thus,the products formed are sulfuric acid $(H_2SO_4)$ and hydrobromic acid $(HBr)$.

(C) The feasibility of displacement reactions between halogens depends on their standard reduction potentials. $A$ halogen with a higher reduction potential can displace a halogen with a lower reduction potential from its salt.
The order of oxidizing power is $F_2 > Cl_2 > Br_2 > I_2$.
In option $(C)$,$I_2$ is a weaker oxidizing agent than $Br_2$. Therefore,$I_2$ cannot displace $Br^-$ from $KBr$ to form $Br_2$. Thus,the reaction $2KBr + I_2 \to 2KI + Br_2$ is not feasible.

(B) When potassium iodide $(KI)$ reacts with copper$(II)$ sulfate $(CuSO_4)$,a redox reaction occurs.
$2CuSO_4 + 4KI \rightarrow 2CuI + I_2 + 2K_2SO_4$
In this reaction,$Cu^{2+}$ ions oxidize $I^-$ ions to iodine $(I_2)$,while $Cu^{2+}$ is reduced to $Cu^+$,forming a white precipitate of copper$(I)$ iodide $(CuI)$.

(D) The reaction of sodium thiosulphate $(Na_2S_2O_3)$ with iodine $(I_2)$ is a redox reaction where iodine acts as an oxidizing agent.
The balanced chemical equation is:
$2Na_2S_2O_3 + I_2 \rightarrow Na_2S_4O_6 + 2NaI$
In this reaction,sodium thiosulphate is oxidized to sodium tetrathionate $(Na_2S_4O_6)$.

(B) When $H_2S$ gas is passed through an acidified $KMnO_4$ solution,$H_2S$ acts as a reducing agent and is oxidized to elemental sulfur $(S)$.
The balanced chemical equation for the reaction is:
$2KMnO_4 + 3H_2SO_4 + 5H_2S \to K_2SO_4 + 2MnSO_4 + 5S + 8H_2O$
Thus,the correct product is $S$.

(C) The correct answer is $(C)$.
Ferrous ions $(Fe^{2+})$ can be oxidized to ferric ions $(Fe^{3+})$ by acidified potassium permanganate $(KMnO_4)$,which causes the purple color of the permanganate solution to discharge (decolorize).
$5Fe^{2+} + MnO_4^- + 8H^+ \rightarrow 5Fe^{3+} + Mn^{2+} + 4H_2O$
Ferric ions $(Fe^{3+})$ are already in their higher oxidation state and do not react with acidified $KMnO_4$,so the purple color remains unchanged.
Thus,acidified $KMnO_4$ acts as a distinguishing reagent.

(D) In the titration of $KMnO_4$ with $Fe^{2+}$ ions,$KMnO_4$ acts as a self-indicator.
As soon as the equivalence point is reached,the addition of one extra drop of $KMnO_4$ imparts a permanent pink color to the solution.
Therefore,no external indicator is required.

(C) The reaction between barium nitrate and sodium sulphate is given by: $Ba(NO_3)_2(aq) + Na_2SO_4(aq) \to BaSO_4(s) + 2NaNO_3(aq)$.
$BaSO_4$ is a white precipitate.
It is insoluble in dilute acids and does not dissolve even upon the addition of a large amount of water.

(C) The reaction $2KBr + I_2 \to 2KI + Br_2$ is not possible.
According to the electrochemical series,$I_2$ is a weaker oxidizing agent than $Br_2$.
Therefore,$I_2$ cannot displace $Br^-$ ions from the solution to form $Br_2$.

(D) In the given redox reaction,$Mn^{2+}$ ions are produced as a product.
These $Mn^{2+}$ ions increase the rate of the reaction,a phenomenon known as autocatalysis.
Therefore,$Mn^{2+}$ acts as an autocatalyst.

(A) The reduction products of $KMnO_4$ depend on the $pH$ of the medium:
$1$. In neutral medium: $KMnO_4$ is reduced to $MnO_2$.
$2$. In alkaline medium: $KMnO_4$ is reduced to $MnO_2$ (or $MnO_4^{2-}$ in strongly alkaline conditions,but $MnO_2$ is the standard product in mildly alkaline/neutral conditions).
$3$. In acidic medium: $KMnO_4$ is reduced to $Mn^{2+}$.
Thus,the products are $MnO_2, MnO_2, Mn^{2+}$.

(C) . $FeSO_4$ is oxidised and $KMnO_4$ is reduced.
In this redox reaction,the oxidation state of $Mn$ changes from $+7$ in $KMnO_4$ to $+2$ in $MnSO_4$ (reduction).
The oxidation state of $Fe$ changes from $+2$ in $FeSO_4$ to $+3$ in $Fe_2(SO_4)_3$ (oxidation).
The balanced chemical equation is:
$2KMnO_4 + 8H_2SO_4 + 10FeSO_4 \to K_2SO_4 + 2MnSO_4 + 5Fe_2(SO_4)_3 + 8H_2O$

(C) The reaction between sodium thiosulphate (hypo) and copper$(II)$ sulphate is as follows:
$Na_2S_2O_3 + CuSO_4 \to NaCuS_2O_3 + Na_2SO_4$
Initially,a yellow precipitate of copper$(I)$ thiosulphate complex $(NaCuS_2O_3)$ is formed,which causes the discharge of the blue colour of the cupric sulphate solution.

(C) When ferrous sulphate $(FeSO_4 \cdot 7H_2O)$ is exposed to air,it loses water of crystallization and undergoes oxidation to form basic ferric sulphate,which is brown in color.
The reaction is: $4FeSO_4 + 2H_2O + O_2 \rightarrow 4Fe(OH)SO_4$.

(D) Acidified $KMnO_4$ acts as a strong oxidizing agent. It reacts with reducing agents to get decolourised. Among the given options,$HBr$ acts as a reducing agent and reduces $Mn(VII)$ to $Mn(II)$,which is colourless. The reaction is: $2KMnO_4 + 10HBr + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 5Br_2 + 8H_2O$.

(C) When iron reacts with dilute $HNO_3$,it is oxidized to ferrous nitrate,while the nitric acid is reduced to ammonium nitrate.
The balanced chemical equation is:
$4Fe + 10HNO_3 \to 4Fe(NO_3)_2 + NH_4NO_3 + 3H_2O$

(C) In acidic medium,$KMnO_4$ acts as a strong oxidising agent and is reduced to $Mn^{2+}$ ions:
$MnO_4^{-} + 8H^{+} + 5e^{-} \to Mn^{2+} + 4H_2O$
In basic (alkaline) medium,$KMnO_4$ is reduced to manganese dioxide $(MnO_2)$:
$MnO_4^{-} + 2H_2O + 3e^{-} \to MnO_2 + 4OH^{-}$
Therefore,the products obtained in acidic and basic media are $Mn^{2+}$ and $MnO_2$ respectively. The correct option is $C$.

(D) When ferrous sulphate $(FeSO_4 \cdot 7H_2O)$ is heated strongly,it first loses water of crystallization and then decomposes to form ferric oxide,sulphur dioxide,and sulphur trioxide.
The balanced chemical equation is:
$2FeSO_4(s) \xrightarrow{\Delta} Fe_2O_3(s) + SO_2(g) + SO_3(g)$

(B) When $AgCl$ is heated with $Na_2CO_3$,it undergoes a decomposition reaction to form metallic silver $(Ag)$,sodium chloride $(NaCl)$,carbon dioxide $(CO_2)$,and oxygen $(O_2)$.
The balanced chemical equation is:
$4AgCl + 2Na_2CO_3 \to 4Ag + 4NaCl + 2CO_2 + O_2$
Thus,the correct product is $Ag$.

(A) When silver nitrate $(AgNO_3)$ is heated to red hot,it undergoes thermal decomposition to form metallic silver,nitrogen dioxide,and oxygen gas.
The balanced chemical equation is:
$2AgNO_3 \xrightarrow{\Delta} 2Ag + 2NO_2 + O_2$
Therefore,the product formed is silver $(Ag)$.

(C) The correct answer is $C$.
Gold is a noble metal and is chemically inert towards common mineral acids like dilute $HNO_3$ when used singly.
Since the copper coin is completely covered with a gold film,the acid cannot come into contact with the copper metal.
Therefore,no chemical reaction occurs,and no nitrate salt is formed.

(C) $HOOC-COOH$ (Oxalic acid) is a reducing agent that reacts with acidic $KMnO_4$ and undergoes oxidation to form $CO_2$ and $H_2O$.
This redox reaction reduces the purple $MnO_4^-$ ion to the colorless $Mn^{2+}$ ion,thereby decolourising the solution.
The balanced chemical equation is:
$2KMnO_4 + 5H_2C_2O_4 + 3H_2SO_4 \rightarrow K_2SO_4 + 2MnSO_4 + 10CO_2 + 8H_2O$

(A) The given reaction $Zn_{(s)} + CuSO_{4(aq)} \rightarrow ZnSO_{4(aq)} + Cu_{(s)}$ is a redox reaction where zinc displaces copper from its salt solution. This reaction occurs naturally and proceeds in the forward direction without external intervention,making it a $Spontaneous \ process$.

(B) In the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$,the oxidation state of $S$ in $SO_2$ changes from $+4$ to $0$ (reduction).
Thus,$SO_2$ acts as the oxidizing agent.
The change in oxidation state per molecule of $SO_2$ is $|4 - 0| = 4$.
The equivalent weight of $SO_2$ is $\frac{\text{Molar Mass}}{n\text{-factor}} = \frac{64}{4} = 16$.
Twice the equivalent weight is $16 \times 2 = 32$.

(C) The balanced redox reaction in acidic medium is:
$5FeC_2O_4 + 3MnO_4^- + 24H^+ \rightarrow 5Fe^{3+} + 10CO_2 + 3Mn^{2+} + 12H_2O$
Alternatively,using the concept of equivalent weights:
$n$-factor of $FeC_2O_4$ (where $Fe^{2+} \rightarrow Fe^{3+} + e^-$ and $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$) is $1 + 2 = 3$.
$n$-factor of $MnO_4^-$ in acidic medium is $5$.
Equating equivalents: $n_{FeC_2O_4} \times (n\text{-factor}) = n_{MnO_4^-} \times (n\text{-factor})$
$1 \times 3 = x \times 5$
$x = \frac{3}{5} = 0.6 \ mol$.

(B) Acidic $KMnO_4$ is a strong oxidizing agent. It gets decolorized by reducing agents.
Among the given options,the sulfide ion $(S^{2-})$ acts as a reducing agent.
The reaction is: $5S^{2-} + 2MnO_4^- + 16H^+ \rightarrow 5S + 2Mn^{2+} + 8H_2O$.
Since $S^{2-}$ is oxidized to $S$,the purple color of $MnO_4^-$ disappears.

(D) $KMnO_4$ is a strong oxidizing agent in an acidic medium.
It reacts with reducing agents like $NO_2^-$,$S^{2-}$,and $Cl^-$ (to some extent),causing the purple color of $KMnO_4$ to disappear or change.
$CO_3^{2-}$ is not a reducing agent and does not react with $KMnO_4$ in an acidic medium,so the color of the $KMnO_4$ solution will not change.

(B) The reaction of sodium thiosulphate $[Na_2S_2O_3]$ with dilute $HCl$ produces sulphur dioxide gas $[SO_2]$ and a yellow precipitate of sulphur $[S]$.
$Na_2S_2O_3 + 2HCl \rightarrow 2NaCl + SO_2 \uparrow + S \downarrow + H_2O$
$(A) = Na_2S_2O_3, (B) = SO_2$
When $SO_2$ gas is passed through acidified potassium dichromate $(K_2Cr_2O_7 + H_2SO_4)$,it reduces $Cr^{6+}$ to $Cr^{3+}$,resulting in a green solution of chromium$(III)$ sulphate $[Cr_2(SO_4)_3]$.
$K_2Cr_2O_7 + H_2SO_4 + 3SO_2 \rightarrow K_2SO_4 + Cr_2(SO_4)_3 + H_2O$
$(C) = Cr_2(SO_4)_3$
Therefore,the correct sequence is $Na_2S_2O_3, SO_2, Cr_2(SO_4)_3$.

(C) In a reaction where the oxidation number does not change,it is a non-redox reaction (e.g.,acid-base neutralization).
In reaction $C$,$Na_2O + H_2SO_4 \to Na_2SO_4 + H_2O$:
Oxidation states of $Na$ is $+1$,$O$ is $-2$,$H$ is $+1$,and $S$ is $+6$ on both sides.
Since there is no change in the oxidation state of any element,this is not a redox reaction.

(A) redox reaction involves a change in the oxidation state of the elements involved.
In the reaction $CaCO_3 \to CaO + CO_2$,the oxidation states are:
$Ca: +2, C: +4, O: -2$ in $CaCO_3$,
$Ca: +2, O: -2$ in $CaO$,
$C: +4, O: -2$ in $CO_2$.
Since there is no change in the oxidation states of any element,this is a thermal decomposition reaction,not a redox reaction.
In other options $(B, C, D)$,the oxidation states of the elements change,indicating they are redox reactions.

(C) When $KI$ is added to $CuSO_4$,$Cu^{2+}$ ions are reduced to $Cu^+$ by $I^-$ ions,and $I^-$ is oxidized to $I_2$.
The balanced chemical equation is:
$2CuSO_4 + 4KI \to Cu_2I_2 + 2K_2SO_4 + I_2$

(C) In the reaction $SO_2 + 2H_2S \to 3S + 2H_2O$,the oxidation state of sulfur in $SO_2$ changes from $+4$ to $0$.
The oxidation state of sulfur in $H_2S$ changes from $-2$ to $0$.
Since $SO_2$ is the oxidizing agent,it undergoes a reduction of $4$ units per molecule ($+4$ to $0$).
The equivalent weight of an oxidizing agent is calculated as $\frac{\text{Molar Mass}}{\text{Change in oxidation number}}$.
Molar mass of $SO_2 = 32 + (2 \times 16) = 64 \ g/mol$.
Equivalent weight $= \frac{64}{4} = 16$.