Redox reaction and Method for balancing Redox reaction Questions in English

Step-$1$: Write the unbalanced equation for the reaction in ionic form:
$Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate the equation into two half-reactions.
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance the atoms other than $O$ and $H$ in each half-reaction. For the reduction half-reaction,multiply $Cr^{+3}$ by $2$ to balance $Cr$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: For reactions in acidic medium,add $H_{2}O$ to balance $O$ atoms and $H^{+}$ to balance $H$ atoms:
$Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$5$: Add electrons to balance the charges. Multiply the oxidation half-reaction by $6$ to equalize the electrons:
Oxidation half: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^{-}$
Reduction half: $Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} + 6e^{-} \rightarrow 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$
Step-$6$: Add the two half-reactions to obtain the balanced overall reaction:
$6Fe_{(aq)}^{+2} + Cr_{2}O_{7_{(aq)}}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_{2}O_{(l)}$

Step-$1$: Write the unbalanced ionic equation:
$Fe_{(aq)}^{+2} + Cr_2O_{7(aq)}^{-2} \rightarrow Fe_{(aq)}^{+3} + Cr_{(aq)}^{+3}$
Step-$2$: Separate into half-reactions:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_2O_{7(aq)}^{-2} \rightarrow Cr_{(aq)}^{+3}$
Step-$3$: Balance atoms other than $O$ and $H$:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3}$
Reduction half: $Cr_2O_{7(aq)}^{-2} \rightarrow 2Cr_{(aq)}^{+3}$
Step-$4$: Balance $O$ atoms using $H_2O$ and $H$ atoms using $H^+$:
Reduction half: $Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} \rightarrow 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$
Step-$5$: Balance charges by adding electrons:
Oxidation half: $Fe_{(aq)}^{+2} \rightarrow Fe_{(aq)}^{+3} + e^-$
Reduction half: $Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} + 6e^- \rightarrow 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$
Step-$6$: Equalize electrons and add the half-reactions:
Multiply oxidation half by $6$: $6Fe_{(aq)}^{+2} \rightarrow 6Fe_{(aq)}^{+3} + 6e^-$
Add both: $6Fe_{(aq)}^{+2} + Cr_2O_{7(aq)}^{-2} + 14H_{(aq)}^{+} \rightarrow 6Fe_{(aq)}^{+3} + 2Cr_{(aq)}^{+3} + 7H_2O_{(l)}$

(N/A) $2MnO_{4(aq)}^- + I_{(aq)}^- + H_2O_{(l)} \to 2MnO_{2(s)} + I_{2(s)} + 2OH_{(aq)}^-$
$(b)$ $2MnO_{4(aq)}^- + 5SO_{2(g)} + 2H_2O_{(l)} \to 2Mn_{(aq)}^{2+} + 5HSO_{4(aq)}^- + H_{(aq)}^+$
$(c)$ $H_2O_{2(aq)} + 2Fe_{(aq)}^{2+} + 2H_{(aq)}^+ \to 2Fe_{(aq)}^{3+} + 2H_2O_{(l)}$
$(d)$ $Cr_2O_7^{2-} + 3SO_{2(g)} + 2H_{(aq)}^+ \to 2Cr_{(aq)}^{3+} + 3SO_{4(aq)}^{2-} + H_2O_{(l)}$

(A) For reaction $(a)$: $P_{4(s)} + 3OH_{(aq)}^{-} + 3H_{2}O_{(l)} \to PH_{3(g)} + 3HPO_{2(aq)}^{-}$
$1$. Oxidation half-reaction: $P_{4} \to 4HPO_{2}^{-} + 8e^{-}$
$2$. Reduction half-reaction: $P_{4} + 12e^{-} \to 4PH_{3}$
$3$. Multiply oxidation half by $3$ and reduction half by $2$ to balance electrons: $3P_{4} + 2P_{4} + 12OH^{-} + 12H_{2}O \to 12HPO_{2}^{-} + 8PH_{3}$.
$4$. Simplified: $5P_{4} + 12OH^{-} + 12H_{2}O \to 8PH_{3} + 12HPO_{2}^{-}$.
Here,$P_{4}$ acts as both the oxidising and reducing agent (disproportionation reaction).

(N/A) The redox reaction involves the oxidation of $SO_2$ to $SO_4^{2-}$ and the reduction of $Cl_2$ to $Cl^-$.
$1.$ Oxidation Half-Reaction $(O.H.R.)$: $SO_{2(aq)} + 2H_2O_{(l)} \rightarrow SO_{4(aq)}^{2-} + 4H^+_{(aq)} + 2e^-$
$2.$ Reduction Half-Reaction $(R.H.R.)$: $Cl_{2(g)} + 2e^- \rightarrow 2Cl^-_{(aq)}$
Adding both half-reactions to cancel the electrons:
$SO_{2(aq)} + Cl_{2(g)} + 2H_2O_{(l)} \rightarrow SO_{4(aq)}^{2-} + 4H^+_{(aq)} + 2Cl^-_{(aq)}$

$(1) H_2S + 2Fe^{3+} \to 2Fe^{2+} + S + 2H^{+}$
$(2) Cu + 2NO_3^{-} + 4H^{+} \to Cu^{2+} + 2NO_2 + 2H_2O$
$(3) 4Sn + NO_3^{-} + 10H^{+} \to 4Sn^{2+} + NH_4^{+} + 3H_2O$
$(4) As + 5NO_3^{-} + 2H_2O \to AsO_4^{3-} + 5NO_2 + 4H^{+}$

(A) When a zinc rod is placed in a $H_2SO_4$ solution,zinc acts as a reducing agent and gets oxidized to $Zn^{2+}$ ions,while hydrogen ions from the acid get reduced to hydrogen gas.
The balanced redox reaction is:
$Zn_{(s)} + H_2SO_{4(aq)} \rightarrow ZnSO_{4(aq)} + H_{2_{(g)}}$
In ionic form,this is represented as:
$Zn_{(s)} + 2H_{(aq)}^{+} \rightarrow Zn_{(aq)}^{2+} + H_{2_{(g)} {2_{(g)}}}$

(C) In the given half-reaction,the manganese atom undergoes reduction.
The oxidation state of $Mn$ in $MnO_4^{-}$ is calculated as: $x + 4(-2) = -1 \implies x = +7$.
The oxidation state of $Mn$ in $Mn^{+2}$ is $+2$.
The change in oxidation state is $|7 - 2| = 5$.
Since the reduction involves a gain of electrons,the number of electrons $n$ required to balance the charge and oxidation state is $5$.
Therefore,$n = 5$.

(N/A) The equivalent weight is calculated as $E = \frac{M}{n}$,where $M$ is the molar mass and $n$ is the change in oxidation state (number of electrons gained).
$1$. Acidic medium: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$. Here,$n = 5$,so equivalent weight is $\frac{M}{5}$.
$2$. Basic medium: $MnO_4^- + 2H_2O + 3e^- \rightarrow MnO_2 + 4OH^-$. Here,$n = 3$,so equivalent weight is $\frac{M}{3}$.
$3$. Neutral or weakly basic medium: $MnO_4^- + e^- \rightarrow MnO_4^{2-}$. Here,$n = 1$,so equivalent weight is $\frac{M}{1}$.

(N/A) An intermolecular redox reaction is a type of reaction where one element in a compound undergoes oxidation while another element in a different compound (or the same compound) undergoes reduction.
Example: $2 KClO_{3} \rightarrow 2 KCl + 3 O_{2}$
In this reaction:
$1$. The oxidation state of $Cl$ changes from $+5$ in $KClO_{3}$ to $-1$ in $KCl$ (Reduction).
$2$. The oxidation state of $O$ changes from $-2$ in $KClO_{3}$ to $0$ in $O_{2}$ (Oxidation).

(8) The balanced chemical equation is: $3Fe + 4H_2O \to Fe_3O_4 + 4H_2$
In the reactant side,the oxidation state of $Fe$ is $0$.
In the product side,$Fe_3O_4$ is a mixed oxide of $FeO$ and $Fe_2O_3$. The oxidation state of $Fe$ in $Fe_3O_4$ is calculated as: $3x + 4(-2) = 0 \implies 3x = 8 \implies x = +8/3$.
Total change in oxidation state for $3$ atoms of $Fe$ is: $3 \times (+8/3 - 0) = 8$.
Therefore,the number of electrons lost by iron is $8$.

(A) True.
In the reaction $Cr_2O_7^{2-} \rightarrow 2Cr^{3+}$,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Total change in oxidation state for two $Cr$ atoms $= 2 \times (6 - 3) = 6$.
Therefore,the n-factor is $6$.
Equivalent mass $= \frac{\text{molecular mass}}{\text{n-factor}} = \frac{\text{molecular mass}}{6}$.

(N/A) In the reaction: $2 F_{2(g)} + 2 H_2O_{(l)} \longrightarrow O_{2(g)} + 4 H^+_{(aq)} + 4 F^-_{(aq)}$
In the above reaction,water acts as a reducing agent; it is oxidized as oxygen changes from an oxidation state of $-2$ in $H_2O$ to $0$ in $O_2$.
Fluorine acts as an oxidizing agent; it is reduced as it changes from an oxidation state of $0$ in $F_2$ to $-1$ in $F^-_{(aq)}$.

(A) complex reaction is one that occurs in more than one elementary step.
In the given reaction,the stoichiometry involves $6$ moles of $FeSO_4$ and $3$ moles of $H_2SO_4$.
It is highly improbable for $10$ molecules $(1 \ KClO_3 + 6 \ FeSO_4 + 3 \ H_2SO_4)$ to collide simultaneously to form products in a single elementary step.
Therefore,the reaction must proceed through a series of elementary steps,making it a complex reaction.
Thus,the statement is True $(T)$.

(B) The reaction between lead$(II)$ sulfide $(PbS)$ and ozone $(O_3)$ is an oxidation reaction.
Ozone acts as a strong oxidizing agent.
The balanced chemical equation is:
$PbS + 4O_3 \,\to PbSO_4 + 4O_2$
Therefore,the missing product is $PbSO_4$.

(A) The reactions are based on the redox properties of the dichromate ion $(Cr_2O_7^{2-})$.
$(a)$ In basic medium,dichromate ions are converted to chromate ions: $Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$. Thus,$X = 2OH^-$.
$(b)$ In acidic medium,dichromate acts as an oxidizing agent: $Cr_2O_7^{2-} + 14H^+ + 6e^- \to 2Cr^{3+} + 7H_2O$. Thus,$X = 6e^-$.
$(c)$ Iodide ions are oxidized to iodine: $6I^- \to 3I_2 + 6e^-$. Thus,$X = 3I_2$.

(N/A) $(1)$ The reaction of dichromate ion with ferrous ion in acidic medium is:
$Cr_2O_7^{2-} + 14H^+ + 6Fe^{2+} \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O$
$(2)$ The reaction of permanganate ion with ferrous ion in acidic medium is:
$MnO_4^- + 8H^+ + 5Fe^{2+} \rightarrow Mn^{2+} + 5Fe^{3+} + 4H_2O$

(N/A) $(1)$ In neutral or faintly alkaline solution,$MnO_4^-$ oxidizes $I^-$ to $IO_3^-$. The balanced equation is:
$2MnO_4^- + I^- + H_2O \rightarrow 2MnO_2 + IO_3^- + 2OH^-$
$(2)$ In acidic medium,$MnO_4^-$ oxidizes oxalate ions $(C_2O_4^{2-})$ to $CO_2$. The balanced equation is:
$2MnO_4^- + 5C_2O_4^{2-} + 16H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$

(N/A) In an acidic medium,$KMnO_4$ acts as a strong oxidizing agent.
When oxalic acid $(H_2C_2O_4)$ is added to an acidic solution of $KMnO_4$,the purple-coloured permanganate ion $(MnO_4^-)$ is reduced to the colourless manganese$(II)$ ion $(Mn^{2+})$.
The balanced chemical equation for this redox reaction is:
$2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$.
Since the product $Mn^{2+}$ is colourless,the characteristic purple colour of the $KMnO_4$ solution disappears.

(N/A) Kolbe's electrolysis is a $redox$ reaction.
At the anode,the carboxylate ion undergoes de-electronation (oxidation) to form an alkyl radical,which then dimerizes to form an alkane.
At the cathode,the water molecules or protons undergo electronation (reduction) to release $H_2$ gas.

(N/A) The unbalanced equation is:
$MnO_{4_{(aq)}}^{-} + SO_{2_{(g)}} \rightarrow Mn_{(aq)}^{+2} + HSO_{4_{(aq)}}^{-}$
Step $1$: Separate into two half-reactions:
Reduction Half-Reaction $(R.H.R.)$ : $MnO_{4_{(aq)}}^{-} \rightarrow Mn_{(aq)}^{+2}$
Oxidation Half-Reaction $(O.H.R.)$ : $SO_{2_{(g)}} \rightarrow HSO_{4_{(aq)}}^{-}$
Step $2$: Balance atoms and charges in each half-reaction in acidic medium:
$R.H.R.$ : $MnO_{4_{(aq)}}^{-} + 8H_{(aq)}^{+} + 5e^{-} \rightarrow Mn_{(aq)}^{+2} + 4H_{2}O_{(l)}$
$O.H.R.$ : $SO_{2_{(g)}} + 2H_{2}O_{(l)} \rightarrow HSO_{4_{(aq)}}^{-} + 3H_{(aq)}^{+} + 2e^{-}$
Step $3$: Equalize the number of electrons by multiplying $R.H.R.$ by $2$ and $O.H.R.$ by $5$:
$2MnO_{4_{(aq)}}^{-} + 16H_{(aq)}^{+} + 10e^{-} \rightarrow 2Mn_{(aq)}^{+2} + 8H_{2}O_{(l)}$
$5SO_{2_{(g)}} + 10H_{2}O_{(l)} \rightarrow 5HSO_{4_{(aq)}}^{-} + 15H_{(aq)}^{+} + 10e^{-}$
Step $4$: Add the two half-reactions:
$2MnO_{4_{(aq)}}^{-} + 5SO_{2_{(g)}} + 16H_{(aq)}^{+} + 10H_{2}O_{(l)}$ $\rightarrow 2Mn_{(aq)}^{+2} + 5HSO_{4_{(aq)}}^{-} + 15H_{(aq)}^{+} + 8H_{2}O_{(l)}$
Simplifying the equation:
$2MnO_{4_{(aq)}}^{-} + 5SO_{2_{(g)}} + 2H_{2}O_{(l)} + H_{(aq)}^{+} \rightarrow 2Mn_{(aq)}^{+2} + 5HSO_{4_{(aq)}}^{-}$

(N/A) Step $1$: Assign oxidation numbers to the atoms undergoing change.
$N_2H_4$: $N$ is $-2$,$H$ is $+1$
$ClO_3^-$: $Cl$ is $+5$,$O$ is $-2$
$NO$: $N$ is $+2$,$O$ is $-2$
$Cl^-$: $Cl$ is $-1$
Step $2$: Identify oxidation and reduction half-reactions.
Oxidation: $N$ changes from $-2$ to $+2$ (loss of $4$ electrons per $N$ atom,total $8$ electrons per $N_2H_4$ molecule).
Reduction: $Cl$ changes from $+5$ to $-1$ (gain of $6$ electrons per $Cl$ atom).
Step $3$: Balance the electrons.
Multiply the oxidation half by $3$ and the reduction half by $4$ to balance the electrons ($24$ electrons transferred).
$3 N_2H_4 + 4 ClO_3^- \rightarrow 6 NO + 4 Cl^-$
Step $4$: Balance $O$ and $H$ atoms in basic medium.
Total $O$ on left: $12$. Total $O$ on right: $6$. Add $6 H_2O$ to the right side.
$3 N_2H_4 + 4 ClO_3^- \rightarrow 6 NO + 4 Cl^- + 6 H_2O$
Check $H$ atoms: Left side has $12 H$ $(3 \times 4)$. Right side has $12 H$ $(6 \times 2)$.
The equation is balanced: $3 N_2H_{4(l)} + 4 ClO_{3(aq)}^- \rightarrow 6 NO_{(g)} + 4 Cl_{(aq)}^- + 6 H_2O_{(l)}$

(N/A) Step $1$: Identify the oxidation and reduction half-reactions.
Reduction Half-Reaction $(R.H.E.)$: $Cl_2O_7(g) \rightarrow ClO_2^-(aq)$
Oxidation Half-Reaction $(O.H.E.)$: $H_2O_2(aq) \rightarrow O_2(g)$
Step $2$: Balance atoms and charges in each half-reaction in an acidic medium.
$R.H.E.$: $Cl_2O_7 + 6H^+ + 8e^- \rightarrow 2ClO_2^- + 3H_2O$
$O.H.E.$: $H_2O_2 \rightarrow O_2 + 2H^+ + 2e^-$
Step $3$: Multiply the $O.H.E.$ by $4$ to equalize the number of electrons $(8e^-)$.
$4H_2O_2 \rightarrow 4O_2 + 8H^+ + 8e^-$
Step $4$: Add the two half-reactions and simplify.
$Cl_2O_7(g) + 6H^+(aq) + 8e^- + 4H_2O_2(aq) \rightarrow 2ClO_2^-(aq) + 3H_2O(l) + 4O_2(g) + 8H^+(aq) + 8e^-$
Final Balanced Equation:
$Cl_2O_7(g) + 4H_2O_2(aq) \rightarrow 2ClO_2^-(aq) + 4O_2(g) + 3H_2O(l) + 2H^+(aq)$

(A) To balance the redox reactions,we use the ion-electron method:
$1. Cr_{2}O_{7}^{2-} + 14H^{+} + 6I^{-} \to 2Cr^{3+} + 3I_{2} + 7H_{2}O$
$2. Cr_{2}O_{7}^{2-} + 6Fe^{2+} + 14H^{+} \to 2Cr^{3+} + 6Fe^{3+} + 7H_{2}O$
$3. 2MnO_{4}^{-} + 5SO_{3}^{2-} + 6H^{+} \to 2Mn^{2+} + 5SO_{4}^{2-} + 3H_{2}O$
$4. 2MnO_{4}^{-} + 16H^{+} + 10Br^{-} \to 2Mn^{2+} + 5Br_{2} + 8H_{2}O$

(D) For the first reaction in basic medium:
$[Fe^{2+} \rightarrow Fe^{3+} + e^{-}] \times 2$
$H_2O_2 + 2 e^{-} \rightarrow 2 OH^{-}$
Adding these gives: $2 Fe^{2+} + H_2O_2 \rightarrow 2 Fe^{3+} + 2 OH^{-}$
Here,$x = 2$ and $y = 2$.
For the second reaction in acidic medium:
$[MnO_4^{-} + 8 H^{+} + 5 e^{-} \rightarrow Mn^{2+} + 4 H_2O] \times 2$
$[H_2O_2 \rightarrow O_2 + 2 H^{+} + 2 e^{-}] \times 5$
Adding these gives: $2 MnO_4^{-} + 16 H^{+} + 5 H_2O_2 \rightarrow 2 Mn^{2+} + 8 H_2O + 5 O_2 + 10 H^{+}$
Simplifying: $2 MnO_4^{-} + 6 H^{+} + 5 H_2O_2 \rightarrow 2 Mn^{2+} + 8 H_2O + 5 O_2$
Here,$x' = 2, y' = 8, z' = 5$.
The sum of stoichiometric coefficients is $x + y + x' + y' + z' = 2 + 2 + 2 + 8 + 5 = 19$.

(C) The balanced chemical equation for the reaction is: $2KMnO_4 + 3H_2SO_4 + 5H_2O_2 \rightarrow K_2SO_4 + 2MnSO_4 + 8H_2O + 5O_2$.
According to the law of equivalence,$Eq$ of $H_2O_2 = Eq$ of $KMnO_4$.
Number of equivalents of $KMnO_4 = \frac{\text{mass}}{\text{equivalent weight}} = \frac{0.316}{158/5} = \frac{0.316 \times 5}{158} = 0.01 \,Eq$.
Since $Eq$ of $H_2O_2 = 0.01$,and the n-factor for $H_2O_2$ is $2$,the moles of $H_2O_2$ are $\frac{0.01}{2} = 0.005 \,mol$.
Mass of pure $H_2O_2 = 0.005 \times 34 = 0.17 \,g$.
Purity of $H_2O_2 = \frac{\text{mass of pure } H_2O_2}{\text{mass of impure } H_2O_2} \times 100 = \frac{0.17}{0.2} \times 100 = 85\%$.

(D) The given redox reaction is: $2 MnO_{4}^{-} + b C_{2}O_{4}^{2-} + c H^{+} \rightarrow x Mn^{2+} + y CO_{2} + z H_{2}O$
Step $1$: Write the reduction half-reaction:
$MnO_{4}^{-} + 8 H^{+} + 5 e^{-} \rightarrow Mn^{2+} + 4 H_{2}O$
Step $2$: Write the oxidation half-reaction:
$C_{2}O_{4}^{2-} \rightarrow 2 CO_{2} + 2 e^{-}$
Step $3$: Equalize the number of electrons by multiplying the reduction half-reaction by $2$ and the oxidation half-reaction by $5$:
$2 MnO_{4}^{-} + 16 H^{+} + 10 e^{-} \rightarrow 2 Mn^{2+} + 8 H_{2}O$
$5 C_{2}O_{4}^{2-} \rightarrow 10 CO_{2} + 10 e^{-}$
Step $4$: Add the two half-reactions:
$2 MnO_{4}^{-} + 5 C_{2}O_{4}^{2-} + 16 H^{+} \rightarrow 2 Mn^{2+} + 10 CO_{2} + 8 H_{2}O$
Comparing this with the given equation,the coefficient $c$ corresponding to $H^{+}$ is $16$.

(C) The reaction of white phosphorus $(P_4)$ with alkali $(NaOH)$ is a disproportionation reaction:
$P_4 + 3 OH^{-} + 3 H_2O \rightarrow PH_3 + 3 H_2PO_2^{-}$
Here,the product '$A$' is the hypophosphite ion $(H_2PO_2^{-})$.
The reaction of $1 \ mol$ of $H_2PO_2^{-}$ with excess $AgNO_3$ is a redox reaction:
$H_2PO_2^{-} + 4 Ag^{+} + 2 H_2O \rightarrow 4 Ag + H_3PO_4 + 3 H^{+}$
From the stoichiometry of the balanced equation,$1 \ mol$ of $H_2PO_2^{-}$ reacts to produce $4 \ mol$ of $Ag$.

(C) The reaction is a disproportionation reaction of $S_8$ in an alkaline medium.
Reduction half-reaction: $S_8 + 16e^{-} \rightarrow 8S^{2-}$
Oxidation half-reaction: $S_8 + 12H_2O \rightarrow 4S_2O_3^{2-} + 24H^{+} + 16e^{-}$
Adding both half-reactions: $2S_8 + 12H_2O \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 24H^{+}$
To balance in basic medium,add $24OH^{-}$ to both sides:
$2S_8 + 12H_2O + 24OH^{-} \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 24H_2O$
Simplifying the equation:
$2S_8 + 24OH^{-} \rightarrow 8S^{2-} + 4S_2O_3^{2-} + 12H_2O$
Dividing by $2$ to get the simplest integer coefficients:
$S_8 + 12OH^{-} \rightarrow 4S^{2-} + 2S_2O_3^{2-} + 6H_2O$
Comparing this with the given equation,the value of '$a$' is $12$.

(C) In mildly alkaline medium,the reaction between thiosulphate ion $(S_2O_3^{2-})$ and permanganate ion $(MnO_4^-)$ is as follows:
$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$
Here,the product $A$ is the sulphate ion $(SO_4^{2-})$.
To find the oxidation state of sulphur $(S)$ in $SO_4^{2-}$:
Let the oxidation state of $S$ be $x$.
$x + 4(-2) = -2$
$x - 8 = -2$
$x = +6$
Therefore,the oxidation state of sulphur in $A$ is $6$.

(D) The aqueous solution of $Fe^{2+}$ ions is green in color.
The aqueous solution of $Fe^{3+}$ ions is yellow in color.
Therefore,the colors of $Fe^{2+}$ and $Fe^{3+}$ ions are green and yellow,respectively.

(B) metal displacement reaction involves a metal in a compound being displaced by another metal in the uncombined state.
In the reaction $Cr_{2}O_{3} + 2 Al \stackrel{\Delta}{\longrightarrow} Al_{2}O_{3} + 2 Cr$,the metal $Al$ displaces the metal $Cr$ from its oxide $Cr_{2}O_{3}$.
Therefore,this is a metal displacement reaction.

(D) In neutral or faintly alkaline medium,the reaction is:
$2KMnO_4 + H_2O + KI \rightarrow 2MnO_2 + 2KOH + KIO_3$
In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
Therefore,the change in oxidation state of manganese is from $+7$ to $+4$.

(A) In acidic medium,the reduction reaction of the permanganate ion is: $MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$.
In $MnO_4^-$,the oxidation state of $Mn$ is $+7$.
In $Mn^{2+}$,the oxidation state of $Mn$ is $+2$.
The change in the oxidation number of manganese is $|(+2) - (+7)| = 5$.

(D) In a neutral or weakly alkaline medium,the permanganate ion $(MnO_{4}^{-})$ acts as an oxidizing agent and gets reduced to manganese dioxide $(MnO_{2})$.
Thiosulphate $(S_{2}O_{3}^{2-})$ is oxidized to sulphate $(SO_{4}^{2-})$.
The balanced chemical equation for this reaction is:
$8 MnO_{4}^{-} + 3 S_{2}O_{3}^{2-} + H_{2}O \rightarrow 8 MnO_{2} + 6 SO_{4}^{2-} + 2 OH^{-}$

(A) In oxalic acid $(H_{2}C_{2}O_{4})$,the oxidation state of carbon is $+3$.
In carbon dioxide $(CO_{2})$,the oxidation state of carbon is $+4$.
The change in oxidation number of carbon is $|4 - 3| = 1$.
The balanced chemical equation is: $2KMnO_{4} + 5H_{2}C_{2}O_{4} + 3H_{2}SO_{4} \rightarrow K_{2}SO_{4} + 2MnSO_{4} + 10CO_{2} + 8H_{2}O$.

(D) The balanced chemical equation for the reaction in a neutral or faintly alkaline medium is:
$8MnO_4^- + 3S_2O_3^{2-} + H_2O \rightarrow 8MnO_2 + 6SO_4^{2-} + 2OH^-$
In $KMnO_4$,the oxidation state of $Mn$ is $+7$.
In $MnO_2$,the oxidation state of $Mn$ is $+4$.
The change in the oxidation state of $Mn$ is $|(+4) - (+7)| = 3$.

(C) The treatment of alkaline $KMnO_{4}$ solution with $KI$ solution oxidizes iodide $(I^{-})$ to iodate $(IO_{3}^{-})$.
The balanced chemical equation is as follows:
$2MnO_{4}^{-} + I^{-} + H_{2}O \longrightarrow 2MnO_{2} + IO_{3}^{-} + 2OH^{-}$

(D) The heating of $(NH_4)_2Cr_2O_7$ produces $Cr_2O_3$ compound along with $N_2$ gas.
$(NH_4)_2Cr_2O_7 \stackrel{\Delta}{\longrightarrow} N_2 + Cr_2O_3 + 4H_2O$
Let the oxidation state of $Cr$ be $x$.
In $(NH_4)_2Cr_2O_7$:
$2(+1) + 2(x) + 7(-2) = 0$
$2 + 2x - 14 = 0$
$2x = 12 \Rightarrow x = +6$
In $Cr_2O_3$:
$2(x) + 3(-2) = 0$
$2x - 6 = 0$
$2x = 6 \Rightarrow x = +3$
Therefore,the change in the oxidation state of $Cr$ in the reaction is from $+6$ to $+3$.

(D) The reaction of hydrogen peroxide $(H_2O_2)$ with potassium permanganate $(KMnO_4)$ in an alkaline medium is given by the equation:
$2KMnO_4 + 3H_2O_2 \rightarrow 2MnO_2 + 2KOH + 2H_2O + 3O_2$
In this reaction,the product containing manganese is manganese dioxide $(MnO_2)$.
To find the oxidation state of $Mn$ in $MnO_2$,let it be $x$.
Since the oxidation state of oxygen is $-2$,we have:
$x + 2(-2) = 0$
$x - 4 = 0$
$x = +4$
Therefore,the oxidation state of $Mn$ is $+4$.

(B) The given unbalanced equation is: $K_2Cr_2O_7 + m \, FeSO_4 + n \, H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + p \, Fe_2(SO_4)_3 + K_2SO_4 + q \, H_2O$
Using the oxidation number method or ion-electron method to balance the redox reaction:
$1$. Oxidation half-reaction: $Fe^{2+} \longrightarrow Fe^{3+} + e^-$
$2$. Reduction half-reaction: $Cr_2O_7^{2-} + 14H^+ + 6e^- \longrightarrow 2Cr^{3+} + 7H_2O$
Multiplying the oxidation half-reaction by $6$ to balance the electrons:
$6Fe^{2+} \longrightarrow 6Fe^{3+} + 6e^-$
Adding the two half-reactions:
$K_2Cr_2O_7 + 6FeSO_4 + 7H_2SO_4 \longrightarrow Cr_2(SO_4)_3 + 3Fe_2(SO_4)_3 + K_2SO_4 + 7H_2O$
Comparing the coefficients,we get $m = 6, n = 7, p = 3, q = 7$.
Thus,the correct option is $B$.

(B) The balanced chemical equation for the oxidation of oxalate ion by $MnO_4^{-}$ in acidic medium is:
$2 MnO_4^{-} + 5 C_2O_4^{2-} + 16 H^{+} \longrightarrow 2 Mn^{2+} + 10 CO_2 + 8 H_2O$
From the stoichiometry,$2$ moles of $MnO_4^{-}$ react with $5$ moles of $C_2O_4^{2-}$.
Thus,the ratio $[MnO_4^{-} : C_2O_4^{2-}] = 2 : 5$.
The balanced chemical equation for the oxidation of $HCl$ by $MnO_4^{-}$ is:
$2 MnO_4^{-} + 16 HCl \longrightarrow 2 MnCl_2 + 5 Cl_2 + 8 H_2O$
From the stoichiometry,$2$ moles of $MnO_4^{-}$ react with $16$ moles of $HCl$.
Thus,the ratio $[MnO_4^{-} : HCl] = 2 : 16 = 1 : 8$.
Therefore,the required ratios are $2:5$ and $1:8$.

(A) The balanced chemical equation for the reaction of $KMnO_4$ with $H_2O_2$ in an acidic medium is:
$2 MnO_4^- + 5 H_2O_2 + 6 H^{+} \longrightarrow 2 Mn^{2+} + 8 H_2O + 5 O_2$
From the stoichiometry of the balanced equation,$2$ moles of $KMnO_4$ produce $5$ moles of $O_2$.
Therefore,the number of moles of $O_2$ produced per mole of $KMnO_4$ is $\frac{5}{2} = 2.5$.

(D) The correct answer is $(iv)$.
$A$ redox reaction is one in which oxidation and reduction occur simultaneously.
$(i) \, CdCl_2 + 2 KOH \longrightarrow Cd(OH)_2 + 2 KCl$: This is a double displacement reaction where no change in oxidation state occurs.
$(ii) \, BaCl_2 + K_2SO_4 \longrightarrow BaSO_4 + 2 KCl$: This is also a double displacement reaction.
$(iii) \, CaCO_3 \longrightarrow CaO + CO_2$: This is a thermal decomposition reaction.
$(iv) \, 2 Ca + O_2 \longrightarrow 2 CaO$: Here,the oxidation state of $Ca$ increases from $0$ to $+2$ (oxidation) and the oxidation state of $O$ decreases from $0$ to $-2$ (reduction). Since both processes occur simultaneously,it is a redox reaction.

(C) The reduction half-reaction for the dichromate ion in an acidic medium is given by:
$Cr_2O_7^{2-} + 14 H^+ + 6 e^- \rightarrow 2 Cr^{3+} + 7 H_2O$
In this reaction,the oxidation state of $Cr$ changes from $+6$ to $+3$.
Since there are two $Cr$ atoms in $Cr_2O_7^{2-}$,the total change in oxidation state is $2 \times (6 - 3) = 6$.
Therefore,$6$ electrons are required to reduce $Cr_2O_7^{2-}$ to $2 Cr^{3+}$.

(D)
When chlorine gas is passed through an aqueous solution of $KBr$,the solution turns orange brown due to the displacement of bromine. Chlorine is a stronger oxidizing agent than bromine,so it displaces bromine from its salt solution.
The balanced chemical equation is:
$Cl_2(g) + 2KBr(aq) \longrightarrow 2KCl(aq) + Br_2(aq/l)$
The formation of $Br_2$ imparts an orange-brown color to the solution.

(B) The reduction half-reaction for permanganate $(MnO_4^-)$ to manganese dioxide $(MnO_2)$ in an acidic medium is given by:
$MnO_4^- + 4H^+ + 3e^- \longrightarrow MnO_2 + 2H_2O$
In this reaction,the oxidation state of manganese changes from $+7$ in $MnO_4^-$ to $+4$ in $MnO_2$.
The change in oxidation state is $7 - 4 = 3$.
Therefore,$3$ electrons are involved in the reduction process.

(D) To balance the chemical equation,the number of atoms of each element must be equal on both sides.
For Iodine $(I)$:
Reactant side: $2$ (from $2 IO_3^-$) $+ x$ (from $x I^-$) $= 2 + x$
Product side: $6 \times 2 = 12$ (from $6 I_2$)
Equating the two sides:
$2 + x = 12$
$x = 12 - 2$
$x = 10$
Thus,the balanced equation is $2 IO_3^- + 10 I^- + 12 H^+ \rightarrow 6 I_2 + 6 H_2 O$.

(C) In faintly alkaline or neutral medium,the permanganate ion $(MnO_4^-)$ is reduced to manganese dioxide $(MnO_2)$.
The balanced half-reaction is:
$MnO_4^- + 2H_2O + 3e^- \longrightarrow MnO_2 + 4OH^-$
From the equation,it is clear that the number of electrons gained is $3$.