Redox reaction and Method for balancing Redox reaction Questions in English

(D) The given reaction is a disproportionation reaction where $P_4$ is both oxidized and reduced.
Step $1$: Write the half-reactions.
Reduction: $P_4 \rightarrow PH_3$
Oxidation: $P_4 \rightarrow H_2 PO_2^-$
Step $2$: Balance the atoms and charges.
Reduction: $P_4 + 3 H_2 O + 3 e^- \rightarrow PH_3 + 3 OH^-$
Oxidation: $P_4 + 8 OH^- \rightarrow 4 H_2 PO_2^- + 4 e^-$
Step $3$: Multiply to equalize electrons and add the equations.
$4(P_4 + 3 H_2 O + 3 e^-$ $\rightarrow PH_3 + 3 OH^-)$ $\Rightarrow 4 P_4 + 12 H_2 O + 12 e^-$ $\rightarrow 4 PH_3 + 12 OH^-$
$3(P_4 + 8 OH^-$ $\rightarrow 4 H_2 PO_2^- + 4 e^-)$ $\Rightarrow 3 P_4 + 24 OH^-$ $\rightarrow 12 H_2 PO_2^- + 12 e^-$
Summing them: $7 P_4 + 12 H_2 O + 12 OH^- \rightarrow 4 PH_3 + 12 H_2 PO_2^-$
Dividing by $4$: $1 P_4 + 3 OH^- + 3 H_2 O \rightarrow 1 PH_3 + 3 H_2 PO_2^-$
Thus,$a=1, b=3, c=1, d=3$.

(A) To balance the redox reaction,we use the half-reaction method:
$1$. Oxidation half-reaction: $C_2O_4^{2-} \rightarrow 2CO_2 + 2e^-$
$2$. Reduction half-reaction: $MnO_4^{-} + 8H^{+} + 5e^- \rightarrow Mn^{2+} + 4H_2O$
$3$. To balance the electrons,multiply the oxidation half-reaction by $5$ and the reduction half-reaction by $2$:
$5C_2O_4^{2-} \rightarrow 10CO_2 + 10e^-$
$2MnO_4^{-} + 16H^{+} + 10e^- \rightarrow 2Mn^{2+} + 8H_2O$
$4$. Adding both half-reactions gives the balanced equation:
$2MnO_4^{-} + 5C_2O_4^{2-} + 16H^{+} \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
Thus,the coefficients for $MnO_4^{-}$,$C_2O_4^{2-}$,and $H^{+}$ are $2$,$5$,and $16$ respectively.

(C) The redox reaction is: $a C_2O_4^{2-} + b MnO_4^- + H^+ \longrightarrow p CO_2 + q Mn^{2+} + H_2O$
Step $1$: Determine the $n$-factors.
For $C_2O_4^{2-}$,the oxidation state of $C$ changes from $+3$ to $+4$. The $n$-factor is $(4-3) \times 2 = 2$.
For $MnO_4^-$,the oxidation state of $Mn$ changes from $+7$ to $+2$. The $n$-factor is $(7-2) \times 1 = 5$.
Step $2$: Balance the $n$-factors by cross-multiplying.
We take $5$ moles of oxalate $(a=5)$ and $2$ moles of permanganate $(b=2)$.
Step $3$: Balance the atoms.
For $C$-atoms: $2a = p \implies p = 2 \times 5 = 10$.
For $Mn$-atoms: $b = q \implies q = 2$.
For $O$-atoms: $5(4) + 2(4) = 10(2) + 2(1) + H_2O \implies 28 = 22 + H_2O \implies H_2O = 6$.
For $H$-atoms: $H^+ = 2 \times 6 = 12$.
The balanced equation is $5C_2O_4^{2-} + 2MnO_4^- + 16H^+ \longrightarrow 10CO_2 + 2Mn^{2+} + 8H_2O$.
Thus,the coefficient of $CO_2$ is $10$.

(A) In the given reaction: $I_2 + 10 HNO_3 \rightarrow 2 HIO_3 + 10 NO_2 + 4 H_2O$.
First,determine the change in the oxidation state of nitrogen in $HNO_3$.
In $HNO_3$,the oxidation state of $N$ is $+5$.
In $NO_2$,the oxidation state of $N$ is $+4$.
The change in oxidation state per molecule of $HNO_3$ is $|5 - 4| = 1$.
Since the change in oxidation state (n-factor) for $HNO_3$ is $1$,the equivalent weight is calculated as:
$\text{Equivalent weight} = \frac{\text{Molar mass}}{\text{n-factor}} = \frac{M}{1} = M$.

(B) The molar mass of $MnO_4^{-}$ is calculated as: $55 + (4 \times 16) = 55 + 64 = 119 \ g/mol$.
In an acidic medium,the reduction reaction is: $MnO_4^{-} + 8H^{+} + 5e^{-} \longrightarrow Mn^{2+} + 4H_2O$.
Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the $n$-factor is $5$.
$\text{Equivalent weight} = \frac{\text{Molar mass}}{n\text{-factor}} = \frac{119}{5} = 23.8$.
Hence,the correct option is $B$.

(D) The balanced chemical equation for the reaction of $MnO_4^{-}$ with $I^{-}$ in acidic medium is:
$2 \ MnO_4^{-} + 10 \ I^{-} + 16 \ H^{+} \longrightarrow 2 \ Mn^{2+} + 5 \ I_2 + 8 \ H_2O$
The $n$-factor for $MnO_4^{-}$ (reduction from $+7$ to $+2$) is $5$.
The $n$-factor for $I^{-}$ (oxidation from $-1$ to $0$ in $I_2$) is $1$ per $I^{-}$ ion.
Using the law of equivalence:
$n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2$
$5 \times 0.02 \times V = 1 \times 0.01 \times 60$
$0.1 \times V = 0.6$
$V = \frac{0.6}{0.1} = 6 \ mL$

(D) The reaction is: $SO_2 + 2H_2O + Cl_2 \longrightarrow H_2SO_4 + 2HCl$
In this reaction,$H_2SO_4$ (sulfuric acid) is a dibasic acid because it has two replaceable hydrogen atoms.
$HCl$ (hydrochloric acid) is a monobasic acid because it has one replaceable hydrogen atom.

(D) According to the law of equivalence,the number of gram equivalents of $H_2 O_2$ must equal the number of gram equivalents of $KMnO_4$.
Gram equivalents $= N \times V \text{ (in Litres)} = M \times n_{factor} \times V \text{ (in Litres)}$.
For $KMnO_4$,the $n_{factor}$ (change in oxidation state of $Mn$ from $+7$ to $+2$) is $5$.
Thus,$N_{H_2 O_2} \times V_{H_2 O_2} = M_{KMnO_4} \times n_{factor} \times V_{KMnO_4}$.
Given: $V_{H_2 O_2} = 20 \ mL$,$M_{KMnO_4} = 0.02 \ M$,$V_{KMnO_4} = 16 \ mL$,$n_{factor} = 5$.
$N_{H_2 O_2} \times 20 = 0.02 \times 5 \times 16$.
$N_{H_2 O_2} = \frac{0.02 \times 5 \times 16}{20} = \frac{0.1 \times 16}{20} = \frac{1.6}{20} = 0.08 \ N$.
$N_{H_2 O_2} = 8 \times 10^{-2} \ N$.

(D) In the presence of a reducing agent,$MnO_4^{-}$ acts as a self-indicator and changes colour from pink to colourless.
Because in $MnO_4^{-}$,the oxidation state of $Mn$ is $+7$,which is its highest oxidation state.
Thus,it tends to get reduced and easily accepts electrons.
Being a charge transfer complex,it shows intense colour,which is why it acts as a self-indicator.

(D) The balanced chemical equation for the reaction in acidic medium is: $10I^{-} + 2MnO_4^{-} + 16H^{+} \rightarrow 5I_2 + 2Mn^{2+} + 8H_2O$.
In $KMnO_4$,the oxidation state of $Mn$ is calculated as: $x + 1(-2) = -1 \Rightarrow x = +7$.
In the product $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
The change in the oxidation number of manganese is $|(+2) - (+7)| = 5$ units.

(A) The redox reaction involves the reduction of $MnO_4^{-}$ to $Mn^{2+}$ and the oxidation of $A^{x+}$ to $AO_3^{-}$.
For $MnO_4^{-} \rightarrow Mn^{2+}$,the change in oxidation state of $Mn$ is from $+7$ to $+2$,which is a gain of $5$ electrons per mole of $MnO_4^{-}$.
For $A^{x+} \rightarrow AO_3^{-}$,the oxidation state of $A$ changes from $+x$ to $+5$. The number of electrons lost per mole of $A$ is $(5 - x)$.
According to the law of equivalence,the total electrons gained equals the total electrons lost.
$1 \times 5 = 1.25 \times (5 - x)$
$5 = 1.25 \times (5 - x)$
$4 = 5 - x$
$x = 5 - 4 = 1$
Therefore,the value of $x$ is $1$.

(A) The balanced redox reaction is:
$NO_3^{-} + 4Mg + 10H^{+} \longrightarrow 4Mg^{2+} + NH_4^{+} + 3H_2O$ (or similar reduction to $NH_3$ in basic medium).
In the reduction of $NO_3^{-}$ to $NH_3$,the oxidation state of $N$ changes from $+5$ to $-3$,so the $n$-factor for $NO_3^{-}$ is $8$.
The oxidation of $Mg$ from $0$ to $+2$ gives an $n$-factor of $2$.
According to the Law of Equivalents:
$Eq. \text{ of } NO_3^{-} = Eq. \text{ of } Mg$
$(n\text{-factor}) \times M \times V(L) = (n\text{-factor}) \times \frac{\text{Mass of } Mg}{\text{Molar mass of } Mg}$
$8 \times 0.1 \times 0.1 = 2 \times \frac{\text{Mass}}{24}$
$0.8 = \frac{\text{Mass}}{12}$
$\text{Mass} = 0.8 \times 1.2 = 0.96 \ g$

(B) The reaction of carbon with concentrated sulfuric acid is an oxidation-reduction reaction.
Carbon is oxidized to carbon dioxide $(CO_2)$ and sulfuric acid is reduced to sulfur dioxide $(SO_2)$.
The balanced chemical equation is:
$C + 2 H_2SO_4 \xrightarrow{\Delta} CO_2 + 2 SO_2 + 2 H_2O$
Thus,$X$ is $CO_2$ and $Y$ is $SO_2$.

(D) The reaction of sodium thiosulphate $(Na_2S_2O_3)$ with moist chlorine $(Cl_2)$ is an oxidation reaction.
In the presence of water,$Na_2S_2O_3$ is oxidized to sodium sulphate $(Na_2SO_4)$ and hydrochloric acid $(HCl)$,while sulphur $(S)$ is precipitated as a byproduct.
The balanced chemical equation is:
$Na_2S_2O_3 + Cl_2 + H_2O \longrightarrow Na_2SO_4 + 2HCl + S$
Here,$X$ corresponds to sulphur $(S)$.

(B) The oxidation of $Fe^{2+}$ to $Fe^{3+}$ involves the loss of $1$ electron: $Fe^{2+} \rightarrow Fe^{3+} + e^-$.
For $MnO_4^-$ in acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Thus,$1$ mole of $MnO_4^-$ oxidizes $5$ moles of $Fe^{2+}$. Given $x = 5$.
For $Cr_2O_7^{2-}$ in acidic medium: $Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O$. Thus,$1$ mole of $Cr_2O_7^{2-}$ oxidizes $6$ moles of $Fe^{2+}$.
Since $x = 5$,the number of moles of $Fe^{2+}$ oxidized by $Cr_2O_7^{2-}$ is $6$. Expressing $6$ in terms of $x$: $6 = \frac{6x}{5}$.

(A) In an acidic medium,the redox reaction between permanganate ions and oxalate ions is given by the balanced equation:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
From the stoichiometry of the balanced equation,$2$ moles of $MnO_4^-$ react with $5$ moles of $C_2O_4^{2-}$.
Therefore,$1$ mole of $MnO_4^-$ reacts with $\frac{5}{2} = 2.5$ moles of $C_2O_4^{2-}$.
Thus,the correct answer is $2.5$.

(D) The given reaction is a disproportionation reaction where $Mn$ in $MnO_4^{2-}$ is oxidized and reduced.
Step $1$: Write the half-reactions.
Oxidation: $MnO_4^{2-} \rightarrow MnO_4^{-} + e^-$
Reduction: $MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^{-}$
Step $2$: Balance the electrons by multiplying the oxidation half-reaction by $2$.
$2MnO_4^{2-} \rightarrow 2MnO_4^{-} + 2e^-$
$MnO_4^{2-} + 2H_2O + 2e^- \rightarrow MnO_2 + 4OH^{-}$
Step $3$: Add the two half-reactions:
$3MnO_4^{2-} + 2H_2O \rightarrow MnO_2 + 2MnO_4^{-} + 4OH^{-}$
Comparing this with the given equation,we get $x=3, y=2, p=1, q=2, r=4$.
Thus,the correct option is $D$.

(C) In acidic medium,$KMnO_4$ (i.e.,$MnO_4^-$) undergoes reduction to $Mn^{2+}$ as follows:
$MnO_4^- + 8H^+ + 5e^- \longrightarrow Mn^{2+} + 4H_2O$
Here,the change in oxidation state of $Mn$ is from $+7$ to $+2$,so the number of electrons gained ($n$-factor) is $5$.
Equivalent weight $= \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{158}{5} = 31.6 \ g$.

(B) In the given reaction in acidic medium,$KMnO_4$ acts as an oxidizing agent.
$KMnO_4$ gains $5$ electrons per molecule to reduce $Mn^{7+}$ to $Mn^{2+}$.
Therefore,the $n$-factor for $KMnO_4$ in acidic medium is $5$.
Equivalent weight = $\frac{\text{Molecular weight}}{n\text{-factor}}$
Equivalent weight = $\frac{158}{5} = 31.6 \ g/\text{equivalent}$.

(D) The balanced redox reaction between potassium permanganate $(KMnO_4)$ and ferrous oxalate $(FeC_2O_4)$ in an acidic medium is:
$3MnO_4^- + 5FeC_2O_4 + 24H^+ \rightarrow 3Mn^{2+} + 5Fe^{3+} + 10CO_2 + 12H_2O$
From the stoichiometry of the balanced equation,$3$ moles of $MnO_4^-$ react with $5$ moles of $FeC_2O_4$.
Therefore,$1$ mole of $MnO_4^-$ will react with $\frac{5}{3}$ moles of $FeC_2O_4$.

(A) $KMnO_4$ oxidises oxalic acid in acidic medium to carbon dioxide. The balanced chemical equation is as follows:
$2 KMnO_4 + 5 H_2 C_2 O_4 + 3 H_2 SO_4 \rightarrow 2 MnSO_4 + 10 CO_2 + 8 H_2 O + K_2 SO_4$
According to the balanced equation,$2 \text{ moles}$ of $KMnO_4$ produce $10 \text{ moles}$ of $CO_2$.
Therefore,the number of $CO_2$ moles produced per mole of $KMnO_4$ is $\frac{10}{2} = 5$.

(D) When dichromate ion oxidises $Fe^{2+}$ ions in aqueous acidic medium to $Fe^{3+}$,the following balanced redox reaction takes place:
$Cr_2O_7^{2-} + 14H^{+} + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
From the balanced equation,it is clear that $14$ protons ($H^{+}$ ions) are consumed for every $1$ mole of dichromate ion reacting with $6$ moles of $Fe^{2+}$ ions.
Therefore,the correct option is $(D)$.

(D) The reduction of nitrogen gas $(N_2)$ to ammonia $(NH_3)$ involves the following balanced half-reaction:
$N_2 + 6e^{-} + 6H^{+} \rightarrow 2NH_3$
In this reaction,the oxidation state of nitrogen changes from $0$ in $N_2$ to $-3$ in $NH_3$.
Since there are two nitrogen atoms,the total change in oxidation state is $2 \times 3 = 6$.
Therefore,$6$ electrons are required for the reduction process.

(C) The reaction between silver nitrate $(AgNO_{3})$ and phosphorous acid $(H_{3}PO_{3})$ is a redox reaction where silver ions are reduced to metallic silver.
The balanced chemical equation is:
$2 \ AgNO_{3} + H_{3}PO_{3} + H_{2}O \longrightarrow H_{3}PO_{4} + 2 \ Ag + 2 \ HNO_{3}$
As shown in the equation,metallic silver $(Ag)$ is produced.

(B) Lime juice contains citric acid,which acts as a source of $H^{+}$ ions. When $H^{+}$ is added to a mixture containing iodide $(I^{-})$ and iodate $(IO_{3}^{-})$ ions,a comproportionation reaction occurs to produce iodine $(I_{2})$,which appears violet in solution. The balanced chemical equation is: $5I^{-} + IO_{3}^{-} + 6H^{+} \longrightarrow 3I_{2} + 3H_{2}O$.

(B) In the given reaction: $I_2 + 2Na_2S_2O_3 \rightarrow 2NaI + Na_2S_4O_6$.
The oxidation state of sulfur in $Na_2S_2O_3$ changes from $+2$ to $+2.5$ in $Na_2S_4O_6$.
The change in oxidation state per sulfur atom is $2.5 - 2 = 0.5$.
Since there are $2$ sulfur atoms in $Na_2S_2O_3$,the total change in oxidation state per molecule is $2 \times 0.5 = 1$.
Thus,the $n$-factor for $Na_2S_2O_3$ is $1$.
Equivalent weight $= \frac{\text{Molecular weight}}{n\text{-factor}} = \frac{M}{1} = M$.

(C) In the given reaction: $2 Cr(OH)_3 + 4 OH^{-} + KIO_3 \rightarrow 2 CrO_4^{2-} + 5 H_2O + KI$
The oxidation state of iodine in $KIO_3$ changes from $+5$ to $-1$ in $KI$.
The change in oxidation state of iodine is $|5 - (-1)| = 6$.
The equivalent weight of an oxidizing agent is given by the formula: $\text{Equivalent weight} = \frac{\text{Molecular mass}}{\text{Change in oxidation state}}$.
Therefore,the equivalent weight of $KIO_3 = M / 6$.

(B) The given reaction is: $2S_2O_3^{2-} + I_2 \longrightarrow S_4O_6^{2-} + 2I^-$
For $S_2O_3^{2-}$: The oxidation state of $S$ changes from $+2$ to $+2.5$. The change in oxidation number per $S$ atom is $0.5$. Since there are $2$ sulfur atoms in $S_2O_3^{2-}$,the total change per mole is $0.5 \times 2 = 1$. Thus,the $n$-factor is $1$. Equivalent weight = $\frac{M_1}{1} = M_1$.
For $I_2$: The oxidation state of $I$ changes from $0$ to $-1$. The change in oxidation number per $I$ atom is $1$. Since there are $2$ iodine atoms in $I_2$,the total change per mole is $1 \times 2 = 2$. Thus,the $n$-factor is $2$. Equivalent weight = $\frac{M_2}{2}$.

(A) The balanced chemical equation is: $2 Na_2S_2O_3 + I_2 \longrightarrow Na_2S_4O_6 + 2 NaI$.
In this reaction,the oxidation state of sulfur in $Na_2S_2O_3$ changes from $+2$ to $+2.5$ in $Na_2S_4O_6$.
The change in oxidation state per sulfur atom is $0.5$. Since there are $2$ sulfur atoms in $Na_2S_2O_3$,the total change in oxidation state (n-factor) is $2 \times 0.5 = 1$.
The equivalent weight $E$ is given by the formula $E = \frac{M}{n\text{-factor}}$,where $M$ is the molar mass.
Thus,$E = \frac{M}{1} = M$.
Therefore,the equivalent weight is equal to the molar mass of sodium thiosulphate.

(B) In an acidic medium,potassium dichromate $(K_{2}Cr_{2}O_{7})$ acts as a strong oxidizing agent.
When it reacts with potassium iodide $(KI)$,the iodide ions $(I^{-})$ are oxidized to iodine $(I_{2})$,and the chromium in the dichromate ion $(Cr_{2}O_{7}^{2-})$ is reduced to the chromic ion $(Cr^{3+})$.
The balanced chemical equation is: $K_{2}Cr_{2}O_{7} + 6KI + 7H_{2}SO_{4} \rightarrow 4K_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + 3I_{2} + 7H_{2}O$.
In the product $Cr_{2}(SO_{4})_{3}$,the oxidation state of chromium is $+3$.

(C) First,balance the redox reaction in acidic medium:
Oxidation half-reaction: $BH_{4}^{-} + 3H_{2}O \rightarrow H_{2}BO_{3}^{-} + 8H^{+} + 8e^{-}$
Reduction half-reaction: $ClO_{3}^{-} + 6H^{+} + 6e^{-} \rightarrow Cl^{-} + 3H_{2}O$
To balance the electrons,multiply the oxidation half-reaction by $3$ and the reduction half-reaction by $4$:
$3BH_{4}^{-} + 9H_{2}O \rightarrow 3H_{2}BO_{3}^{-} + 24H^{+} + 24e^{-}$
$4ClO_{3}^{-} + 24H^{+} + 24e^{-} \rightarrow 4Cl^{-} + 12H_{2}O$
Adding these,we get the balanced equation: $3BH_{4}^{-} + 4ClO_{3}^{-} \rightarrow 3H_{2}BO_{3}^{-} + 4Cl^{-} + 3H_{2}O$
In the Nernst equation,$n$ represents the number of moles of electrons transferred in the balanced redox reaction.
From the balanced equation,$n = 24$.