Auto oxidation and Disproportionation Questions in English

(C) The reaction is: $P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$.
In this reaction,the oxidation state of phosphorus $(P)$ changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).
Since both oxidation and reduction occur simultaneously,it is a disproportionation reaction,which is a type of redox reaction.

(A) In the reaction $2CuI \to Cu + CuI_2$,the oxidation state of $Cu$ in $CuI$ is $+1$.
In the product side,$Cu$ is in $0$ oxidation state and $Cu$ in $CuI_2$ is in $+2$ oxidation state.
Since $Cu$ is simultaneously oxidized (from $+1$ to $+2$) and reduced (from $+1$ to $0$),this is a disproportionation reaction,which is a type of redox reaction.

(D) In the given reaction: $3Br_2^0 + 6CO_3^{2-} + 3H_2O \rightarrow 5Br^{-1} + Br^{+5}O_3^- + 6HCO_3^-$.
The oxidation state of $Br$ changes from $0$ to $-1$ (reduction) and from $0$ to $+5$ (oxidation).
Therefore,bromine undergoes both oxidation and reduction,which is a disproportionation reaction.

(A) In the reaction $4P + 3KOH + 3H_2O \to 3KH_2PO_2 + PH_3$,the oxidation state of $P$ changes as follows:
$1$. In elemental phosphorus $(P_4)$,the oxidation state of $P$ is $0$.
$2$. In $KH_2PO_2$,the oxidation state of $P$ is calculated as: $1 + 2(+1) + x + 2(-2) = 0 \implies 1 + 2 + x - 4 = 0 \implies x = +1$.
$3$. In $PH_3$,the oxidation state of $P$ is calculated as: $x + 3(+1) = 0 \implies x = -3$.
Since the oxidation state of $P$ increases from $0$ to $+1$ (oxidation) and decreases from $0$ to $-3$ (reduction),$P$ is both oxidized and reduced. This is a disproportionation reaction.

(B) In $SO_2$,sulphur has an oxidation state of $+4$. Since the maximum oxidation state of sulphur is $+6$ and the minimum is $-2$,it can be oxidised to $+6$ or reduced to $-2$. Thus,$SO_2$ can act as both an oxidising and a reducing agent.
In $H_2SO_4$,sulphur is in its highest oxidation state of $+6$,so it can only act as an oxidising agent.
In $H_2S$,sulphur is in its lowest oxidation state of $-2$,so it can only act as a reducing agent.
In $HNO_3$,nitrogen is in its highest oxidation state of $+5$,so it can only act as an oxidising agent.

(D) In the reaction $H_2O + Br_2 \to HOBr + HBr$,the oxidation state of bromine changes as follows:
$Br_2$ (elemental state) has an oxidation number of $0$.
In $HOBr$,the oxidation number of $Br$ is $+1$.
In $HBr$,the oxidation number of $Br$ is $-1$.
Since the oxidation number of bromine increases from $0$ to $+1$ (oxidation) and decreases from $0$ to $-1$ (reduction) simultaneously,bromine undergoes both oxidation and reduction.
Therefore,it is a disproportionation reaction,which is a type of redox reaction.

(D) $NaNO_2$ (Sodium nitrite) acts as both an oxidising and a reducing agent because the $N$ atom is in the $+3$ oxidation state,which is an intermediate oxidation state between its minimum $(-3)$ and maximum $(+5)$ values.
Oxidising property:
$2NaNO_2 + 2KI + 2H_2SO_4 \xrightarrow{} Na_2SO_4 + K_2SO_4 + 2NO + 2H_2O + I_2$
Reducing property:
$H_2O_2 + NaNO_2 \xrightarrow{} NaNO_3 + H_2O$

(C) In the reaction $P + NaOH \to PH_3 + NaH_2PO_2$,the oxidation state of phosphorus $(P)$ changes from $0$ in $P$ to $-3$ in $PH_3$ and $+1$ in $NaH_2PO_2$.
Since the oxidation state of $P$ both decreases and increases,it undergoes both reduction and oxidation.
This is a disproportionation reaction where $P$ is simultaneously oxidized and reduced.

(D) The correct answer is $D$.
$HNO_2$ (nitrous acid) contains nitrogen in the $+3$ oxidation state.
Since the oxidation state of nitrogen can range from $-3$ to $+5$,$HNO_2$ can act as an oxidizing agent (by getting reduced) and as a reducing agent (by getting oxidized).
Additionally,the nitrite ion $(NO_2^-)$ acts as an ambidentate ligand,allowing it to form stable coordination complexes with various metal ions.

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in a neutral aqueous medium is given by:
$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$
From the stoichiometry of the balanced equation,$3 \ moles$ of $MnO_4^{2-}$ produce $2 \ moles$ of $MnO_4^-$ and $1 \ mole$ of $MnO_2$.
Therefore,$1 \ mole$ of $MnO_4^{2-}$ will produce $\frac{2}{3} \ mole$ of $MnO_4^-$ and $\frac{1}{3} \ mole$ of $MnO_2$.

(C) The thermal decomposition of ammonium dichromate is a classic reaction used in the 'volcano' experiment.
When heated,it decomposes to produce chromium$(III)$ oxide,nitrogen gas,and water vapor.
The balanced chemical equation is:
$(NH_4)_2Cr_2O_7 \xrightarrow{\Delta} Cr_2O_3 + N_2 + 4H_2O$
Therefore,the products are chromium oxide and nitrogen.

(C) The chemical reaction is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction,which is a type of redox reaction involving both oxidation and reduction.

(B) In $P_4$,the oxidation state of phosphorus is $0$.
In $PH_3$,the oxidation state of phosphorus is $-3$ (reduction).
In $NaH_2PO_2$,the oxidation state of phosphorus is $+1$ (oxidation).
Since phosphorus is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) When $Cl_2$ gas reacts with a hot and concentrated solution of $KOH$,it undergoes a disproportionation reaction.
The balanced chemical equation is:
$6KOH + 3Cl_2 \to 5KCl + KClO_3 + 3H_2O$
Thus,potassium chlorate $(KClO_3)$ is formed as the main product.

(C) In this reaction,phosphorus $(P_4)$ is simultaneously oxidised to $NaH_2PO_2$ and reduced to $PH_3$.
The oxidation state of $P$ changes from $0$ in $P_4$ to $+1$ in $NaH_2PO_2$ and $-3$ in $PH_3$.
$\overset{0}{P}_4 + 3NaOH + 3H_2O \to 3NaH_2\overset{+1}{P}O_2 + overset{-3}{P}H_3$
Since the same element is both oxidised and reduced,this is an example of a disproportionation reaction.

(C) When $H_2S$ gas is passed through a solution containing mercurous ions $(Hg_2^{2+})$,a disproportionation reaction occurs.
The reaction is: $Hg_2^{2+} + H_2S \longrightarrow HgS + Hg + 2H^+$.
In this reaction,$Hg_2^{2+}$ is simultaneously oxidized to $Hg^{2+}$ (which forms $HgS$) and reduced to metallic $Hg$.

(B) When $H_2S$ gas is passed through a solution containing $Hg_2^{2+}$ ions,a disproportionation reaction occurs.
The reaction is: $Hg_2^{2+} (aq) + H_2S (g) \rightarrow HgS (s) + Hg (l) + 2H^+ (aq)$.
Thus,a mixture of black mercuric sulfide $(HgS)$ and metallic mercury $(Hg)$ is obtained.

(C) In the reaction: $H_2O + \overset{0}{Br_2} \to HO\overset{+1}{Br} + H\overset{-1}{Br}$
$1$. The oxidation state of $Br$ in $Br_2$ is $0$.
$2$. In $HOBr$,the oxidation state of $Br$ is $+1$ (Oxidation).
$3$. In $HBr$,the oxidation state of $Br$ is $-1$ (Reduction).
Since $Br_2$ undergoes both oxidation and reduction,it is a disproportionation reaction.

(C) The given reaction is: ${P_4} + 3NaOH + 3{H_2}O \to 3Na{H_2}P{O_2} + P{H_3}$.
In this reaction,the oxidation state of phosphorus changes as follows:
${P_4}^0 \to Na{H_2}P^{+1}{O_2}$ (Oxidation,as oxidation state increases from $0$ to $+1$).
${P_4}^0 \to P^{-3}{H_3}$ (Reduction,as oxidation state decreases from $0$ to $-3$).
Since both oxidation and reduction occur simultaneously,this is a disproportionation reaction (a type of redox reaction).

(C) Assigning oxidation states: $3\overset{0}{I}_2 + 6NaOH \to Na\overset{+5}{I}O_3 + 5Na\overset{-1}{I} + 3H_2O$.
In this reaction,the oxidation state of iodine $(I)$ changes from $0$ to $+5$ (oxidation) and from $0$ to $-1$ (reduction).
Since $I_2$ undergoes both oxidation and reduction (disproportionation reaction),it acts as both the oxidizing agent and the reducing agent.
Therefore,$I_2$ is the oxidizing agent.

(D) Assigning oxidation numbers to the species in the reaction:
$3\overset{0}{Br}_2 + 6CO_3^{2-} + 3H_2O \to 5\overset{-1}{Br}^- + \overset{+5}{Br}O_3^- + 6HCO_3^-$
$1$. The oxidation state of $Br$ changes from $0$ in $Br_2$ to $-1$ in $Br^-$,which is a decrease in oxidation number,representing reduction.
$2$. The oxidation state of $Br$ changes from $0$ in $Br_2$ to $+5$ in $BrO_3^-$,which is an increase in oxidation number,representing oxidation.
Since $Br_2$ undergoes both oxidation and reduction,it is a disproportionation reaction.

(A) The disproportionation reaction of $MnO_4^{2-}$ is given by:
$3MnO_4^{2-} + 2H_2O \rightarrow 2MnO_4^- + MnO_2 + 4OH^-$
In this reaction,$3 \ moles$ of $MnO_4^{2-}$ produce $2 \ moles$ of $MnO_4^-$ and $1 \ mole$ of $MnO_2$.
Therefore,for $1 \ mole$ of $MnO_4^{2-}$,the products are:
$\frac{2}{3} \ mol$ of $MnO_4^-$ and $\frac{1}{3} \ mol$ of $MnO_2$.

(C) The reaction described is a disproportionation reaction,specifically the Cannizzaro reaction for aldehydes lacking $\alpha$-hydrogen atoms.
$2HCHO + OH^- \to CH_3OH + HCOO^-$
In this reaction,one molecule of formaldehyde $(HCHO)$ is reduced to methanol $(CH_3OH)$ and another molecule is oxidized to formate $(HCOO^-)$.
Therefore,formaldehyde undergoes self-oxidation and self-reduction.

(C) The reaction is: $P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$.
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $+1$ in $NaH_2PO_2$ (oxidation) and to $-3$ in $PH_3$ (reduction).
Since the same element is simultaneously oxidized and reduced,it is a disproportionation reaction.

(C) In the reaction $3\mathop{Cl}\limits^{+1}O{(aq)}^{-} \to \mathop{Cl}\limits^{+5}O_{3(aq)}^{-} + 2\mathop{Cl}\limits^{-1}{(aq)}^{-}$,
the oxidation state of chlorine changes from $+1$ to $+5$ (oxidation) and from $+1$ to $-1$ (reduction).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction.

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ in neutral or acidic medium is given by:
$3MnO_4^{2-} + 2H_2O \to MnO_2 + 2MnO_4^- + 4OH^-$.
To find the products for $1 \ mol$ of $MnO_4^{2-}$,divide the entire equation by $3$:
$MnO_4^{2-} + \frac{2}{3}H_2O \to \frac{1}{3}MnO_2 + \frac{2}{3}MnO_4^- + \frac{4}{3}OH^-$.
Thus,$1 \ mol$ of $MnO_4^{2-}$ produces $\frac{2}{3} \ mol$ of $MnO_4^-$ and $\frac{1}{3} \ mol$ of $MnO_2$.

(D) In $NaNO_2$ (Sodium nitrite),the oxidation state of nitrogen $(N)$ is $+3$.
Nitrogen can be oxidized to a higher oxidation state (e.g.,$+5$ in $NO_3^-$) or reduced to a lower oxidation state (e.g.,$+2, +1, 0, -3$).
Therefore,$NaNO_2$ can act as both an oxidizing agent and a reducing agent.
In $NaNO_3$,$N$ is in its maximum oxidation state $(+5)$,so it acts only as an oxidizing agent.
In $Na_2O$,$O$ is in its minimum oxidation state $(-2)$,so it acts only as a reducing agent.
In $Na_2O_2$,$O$ is in $-1$ state,which can be oxidized to $0$ or reduced to $-2$,but $NaNO_2$ is the classic example of a substance showing dual behavior due to the intermediate oxidation state of nitrogen.

(D) The chemical reaction is: $3Cl_2 (0) + 6NaOH \rightarrow 5NaCl (-1) + NaClO_3 (+5) + 3H_2O$.
Since the oxidation state of chlorine changes from $0$ to $-1$ (reduction) and $0$ to $+5$ (oxidation) simultaneously,it is a redox reaction.

(C) The $Cu^{+}$ ion is unstable in an aqueous solution and undergoes disproportionation reaction as follows:
$2Cu^{+} (aq) \rightarrow Cu (s) + Cu^{+2} (aq)$
In this reaction,$Cu^{+}$ is simultaneously oxidized to $Cu^{+2}$ and reduced to $Cu$ metal.

(D) In the reaction $2Cu^{+} \rightarrow Cu + Cu^{2+}$,the oxidation state of copper changes from $+1$ to $0$ (reduction) and from $+1$ to $+2$ (oxidation).
Since the same species is simultaneously oxidized and reduced,this type of reaction is known as a disproportionation reaction.

(B) The reaction of chlorine gas with hot and concentrated sodium hydroxide solution is:
$3Cl_2 + 6NaOH \rightarrow NaClO_3 + 5NaCl + 3H_2O$
In $Cl_2$,the oxidation number of $Cl$ is $0$.
In $NaCl$,the oxidation number of $Cl$ is $-1$.
In $NaClO_3$,the oxidation number of $Cl$ is $x + (-2 \times 3) = -1 \Rightarrow x = +5$.
Thus,the oxidation number changes from $0$ to $-1$ and $0$ to $+5$.

(B) Hypochlorous acid $(HOCl)$ undergoes disproportionation reaction upon heating or standing.
The reaction is given by:
$3 HOCl \rightarrow 2 HCl + HClO_3$
In this reaction,the chlorine atom in $HOCl$ (oxidation state $+1$) is simultaneously oxidized to $HClO_3$ (oxidation state $+5$) and reduced to $HCl$ (oxidation state $-1$).

(D) The disproportionation reaction is: $3H_3PO_2 \rightarrow PH_3 + 2H_3PO_3$.
In this reaction,the oxidation state of $P$ in $H_3PO_2$ is $+1$.
It is oxidized to $H_3PO_3$ (where $P$ is $+3$) and reduced to $PH_3$ (where $P$ is $-3$).
Change in oxidation state for oxidation: $3 - 1 = 2$.
Change in oxidation state for reduction: $1 - (-3) = 4$.
For a disproportionation reaction,the equivalent weight $E$ is given by $E = E_{ox} + E_{red} = M/n_{ox} + M/n_{red}$.
Here,$n_{ox} = 2$ and $n_{red} = 4$.
Therefore,$E = M/2 + M/4 = (2M + M)/4 = 3M/4$.

(C) The given reaction is the $Cannizzaro$ reaction,where benzaldehyde $(PhCHO)$ undergoes self-oxidation and reduction in the presence of a concentrated base $(OH^-)$.
In this reaction,one molecule of benzaldehyde is reduced to benzyl alcohol $(PhCH_2OH)$ and another molecule is oxidized to benzoate ion $(PhCOO^-)$.
Since both oxidation and reduction occur simultaneously in the same species,it is a disproportionation reaction.
Since it involves both oxidation and reduction,it is also a redox reaction.
Therefore,the reaction is both a redox reaction and a disproportionation reaction.

(A) The disproportionation reaction of manganate ion $(MnO_4^{2-})$ is given by:
$3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$
From the stoichiometry of the balanced equation,$3 \text{ moles}$ of $MnO_4^{2-}$ produce $2 \text{ moles}$ of $MnO_4^-$.
Therefore,$1 \text{ mole}$ of $MnO_4^{2-}$ produces $2/3 \text{ moles}$ of $MnO_4^-$.
The mass of $Mn$ in $2/3 \text{ moles}$ of $MnO_4^-$ is $(2/3) \times 55 \ g$.
The total mass of $Mn$ in $1 \text{ mole}$ of $MnO_4^{2-}$ is $1 \times 55 \ g$.
The percentage by mass of $Mn$ converted to $MnO_4^-$ is:
$\frac{(2/3) \times 55}{55} \times 100 = \frac{2}{3} \times 100 = 66.67\%$.

(C) The reaction is: $3 H_3PO_2 \xrightarrow{\Delta} 3 H_3PO_4 + PH_3$.
In this reaction,the oxidation state of phosphorus in $H_3PO_2$ is $+1$.
In the products,the oxidation state of phosphorus in $H_3PO_4$ is $+5$ and in $PH_3$ is $-3$.
Since the same element $(P)$ is simultaneously oxidized and reduced,this is a disproportionation reaction.

(B) $Cl_2$,$MnO_4^{2-}$,and $NO_2$ undergo disproportionation in alkaline medium.
$1. \, Cl_2 + 2 OH^- \rightarrow Cl^- + ClO^- + H_2 O$
$2. \, 3 MnO_4^{2-} + 4 H^+ \rightarrow 2 MnO_4^- + MnO_2 + 2 H_2 O$ (Note: $MnO_4^{2-}$ disproportionates in neutral or acidic medium,but is often cited in context of its instability relative to $MnO_4^-$).
$3. \, 2 NO_2 + 2 OH^- \rightarrow NO_2^- + NO_3^- + H_2 O$
$ClO_4^-$ does not undergo disproportionation as $Cl$ is already in its highest oxidation state of $+7$.

(B) The chemical equation for the reaction is: $P_4 + 3NaOH + 3H_2O \to PH_3 + 3NaH_2PO_2$
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and $+1$ in $NaH_2PO_2$ (oxidation).
Since the same element is simultaneously oxidized and reduced,this is a disproportionation reaction.

(D) Bleaching powder $(CaOCl_2)$ undergoes auto-oxidation when heated or stored for a long time.
The chemical reaction is: $6CaOCl_2 \rightarrow 5CaCl_2 + Ca(ClO_3)_2$.
Thus,the products formed are calcium chloride $(CaCl_2)$ and calcium chlorate $(Ca(ClO_3)_2)$.

(C) The reaction is: $3 H_3PO_2 \xrightarrow{\Delta} 2 H_3PO_3 + PH_3$ and $4 H_3PO_2 \xrightarrow{\Delta} 3 H_3PO_4 + PH_3$.
In this reaction,the oxidation state of phosphorus in $H_3PO_2$ is $+1$.
In $PH_3$,the oxidation state is $-3$ (reduction),and in $H_3PO_3$ or $H_3PO_4$,the oxidation state is $+3$ or $+5$ respectively (oxidation).
Since the same element is simultaneously oxidized and reduced,it is a disproportionation reaction.

(C) disproportionation reaction is a redox reaction in which the same element is simultaneously oxidized and reduced.
In the reaction $2CuBr \longrightarrow CuBr_2 + Cu$,the copper ion $Cu^{+}$ in $CuBr$ undergoes disproportionation.
The oxidation state of $Cu$ changes from $+1$ in $CuBr$ to $+2$ in $CuBr_2$ (oxidation) and to $0$ in $Cu$ (reduction).
Thus,the reaction is $2Cu^{+} \longrightarrow Cu^{2+} + Cu$.

(B) In aqueous solution,$Cu^{+}$ ions are unstable because they undergo disproportionation reaction:
$2Cu^{+}_{(aq)} \to Cu^{2+}_{(aq)} + Cu_{(s)}$
This occurs because the hydration energy of $Cu^{2+}$ is much higher than that of $Cu^{+}$,which compensates for the energy required to remove the second electron from $Cu^{+}$.

(C) In the given reaction: $P_4 + 3NaOH + 3H_2O \to 3NaH_2PO_2 + PH_3$
The oxidation state of phosphorus $(P)$ in $P_4$ is $0$.
In $NaH_2PO_2$,the oxidation state of $P$ is $+1$ (Oxidation).
In $PH_3$,the oxidation state of $P$ is $-3$ (Reduction).
Since the same element $(P)$ is simultaneously oxidized and reduced in the reaction,it is an example of a disproportionation redox reaction.

(D) In the given reaction,the oxidation state of bromine $(Br_2)$ in the reactant side is $0$.
When it changes to $BrO_3^-$,the oxidation state of bromine increases to $+5$,which indicates oxidation (loss of electrons).
When it changes to $Br^-$,the oxidation state of bromine decreases to $-1$,which indicates reduction (gain of electrons).
Since bromine undergoes both oxidation and reduction,this is a disproportionation reaction.
Therefore,bromine is both reduced and oxidized.

(C) An ion can undergo both oxidation and reduction if the central atom is in an intermediate oxidation state.
In $OCl^-$,the oxidation state of $Cl$ is $+1$.
$Cl$ can be oxidized to a higher state (e.g.,$+3, +5, +7$) or reduced to a lower state (e.g.,$-1$ in $Cl^-$).
In $Cr_2O_7^{2-}$,$Cr$ is in its maximum oxidation state of $+6$,so it can only be reduced.
In $NO_3^-$,$N$ is in its maximum oxidation state of $+5$,so it can only be reduced.
In $S^{2-}$,$S$ is in its minimum oxidation state of $-2$,so it can only be oxidized.

(D) The auto-oxidation of bleaching powder $(CaOCl_2)$ occurs when it is heated or left for a long time,resulting in the formation of calcium chlorate and calcium chloride.
The chemical reaction is: $6CaOCl_2 \rightarrow Ca(ClO_3)_2 + 5CaCl_2$.

(D) The $Bessemer$ process,$Open-Hearth$ process,and $Kaldo$ process are all industrial methods used in the production of steel.
$Auto-oxidation$ is a chemical reaction involving the spontaneous oxidation of a substance in the presence of oxygen,which is not a specific process used in steel manufacturing.
Therefore,$Auto-oxidation$ is the correct answer.

(C) Assigning oxidation states to phosphorus in the given reaction:
$\mathop{P_4}\limits^{0} + 3KOH + 3H_2O \to \mathop{PH_3}\limits^{-3} + 3KH_2\mathop{P}\limits^{+1}O_2$
In this reaction,the oxidation state of phosphorus changes from $0$ to $-3$ (reduction) and from $0$ to $+1$ (oxidation).
Therefore,phosphorus undergoes both oxidation and reduction,which is a disproportionation reaction.

(B) In $HNO_2$,the oxidation state of nitrogen is $+3$. Nitrogen can be oxidized to $+5$ (in $HNO_3$) and reduced to $+2$ (in $NO$).
Therefore,$HNO_2$ undergoes disproportionation reaction:
$3HNO_2 \rightarrow HNO_3 + 2NO + H_2O$
Since it can both increase and decrease its oxidation state,it acts as both an oxidizing and a reducing agent.

(C) The chemical reaction is: $P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$.
In this reaction,the oxidation state of phosphorus changes from $0$ in $P_4$ to $-3$ in $PH_3$ (reduction) and to $+1$ in $NaH_2PO_2$ (oxidation).
Since both oxidation and reduction occur simultaneously,this is a disproportionation reaction,which is a type of redox reaction.