MCQ Questions for Class 11 Medical Chemistry Redox Reactions Quiz 5

Oxidation state of phosphorous in

$\displaystyle \ { P } { O }_{ 4 }^{ 3- }$ is:

0%
+2
0%
+5
0%
+4
0%
+1
0%
none of the above

Explanation

Oxidation state of phosphorous in

$\displaystyle \displaystyle \ { P } { O }_{ 4 }^{ 3- }$ is +5.
Let,

$X$ be the oxidation state of phosphorous.

$\displaystyle X + 4(-2) = -3 \\ X - 8 = -3 \\ X = +5$

$Cl_2(g)+2Br^-(excess)\rightarrow$
What are the final product of the above reaction equation?

0%
$1$ mol of $Cl^-$ ions and $1$ mol of $Br^-$
0%
$1$ mol of $Cl^-$ ions and $2$ mol of $Br^-$
0%
$1$ mol of $Cl^-$ ions and $1$ mol of $Br_2$
0%
$2$ mol of $Cl^-$ ions and $1$ mol of $Br_2$
0%
$2$ mol of $Cl^-$ ions and $2$ mol of $Br_2$

Explanation

The given reaction may be expressed as

$Cl_2(g)+2Br^-(excess)\rightarrow 2Cl^-+Br_2(g)$

Hence, 2 moles of

$Cl^-$ and one mole of

$Br_2$ will be formed

The oxidation number of sulfur in

$H_2SO_4$ is :

0%
$+2$
0%
$+3$
0%
$+4$
0%
$+6$
0%
$+8$

Explanation

Let X be the oxidation number of S in

$\displaystyle H_2SO_4$ .
The oxidation number of hydrogen is +1 and the oxidation number of oxygen is -2.

$\displaystyle 2(+1) +X +4(-2) = 0 \\ 2 +X-8=0 \\ X-6=0 \\ X= +6$
Hence, the oxidation number of S in

$\displaystyle H_2SO_4$ is +6.

Consider

$\displaystyle { Mg\left( { H } { CO }_{ 3 } \right) }_{ 2 }$ and choose the correct option.

0%
Oxidation state of hydrogen is +5
0%
Oxidation state of carbon is +1
0%
Oxidation state of hydrogen is +1
0%
Oxidation state of hydrogen is +4
0%
None of the above

Explanation

The oxidation state of carbon is +4.
The oxidation state of hydrogen is +1.
The oxidation state of oxygen is -2.
The oxidation state of magnesium is +2.
Hence, option

$(C)$ is the correct statement.

What is the oxidation state of sulphur in

$\displaystyle { Na }_{ 2 }\ { { S }_{ 2 } } { O }_{ 3 }$ ?

0%
+2
0%
+4
0%
+1
0%
+5
0%
+6

Explanation

The oxidation state of sulphur in

$\displaystyle \displaystyle { Na }_{ 2 }\ { { S }_{ 2 } } { O }_{ 3 }$ is +2.
Let,

$X$ be the oxidation state of sulphur.

$\displaystyle 2 (+1) + 2 (X) +3(-2) = 0 \\ 2 + 2X - 6 = 0 \\ 2X = 4 \\ X = +2$

Zn(s)|ZnCl

$_2$ (aq)|Cl

$^-$ (aq)|Cl

$_2$ (g)|C(s)
According to the above cell diagram the electrochemical cell described, the reaction at anode is :

0%
Zn $\rightarrow$ Zn $^{2+}$ + 2e $^-$
0%
Zn $^{2+}$ + 2e $^-$ $\rightarrow$ Zn
0%
Cl $_2$ + 2e $^-$ $\rightarrow$ 2Cl $^-$
0%
2Cl $^-$ $\rightarrow$ Cl $_2$ + 2e $^-$
0%
Zn + Cl $_2$ $\rightarrow$ ZnCl $_2$

Explanation

At anode, zinc is oxidized to Zn(II) ions.

$\displaystyle Zn \rightarrow Zn^{2+} + 2e^-$
In the cell diagram, the electrode on left hand side is anode and the electrode on right hand side is cathode. Oxidation occurs at anode and reduction occurs at cathode.

Oxidation state for the underlined atom in

$\displaystyle { K }_{ 2 }\underline { Cr } { O }_{ 4 }$ is:

0%
+4
0%
+6
0%
+2
0%
+5
0%
+1

Explanation

Oxidation state for the underlined atom in

$\displaystyle \displaystyle { K }_{ 2 }\underline { Cr } { O }_{ 4 }$ is +6.
Let,

$X$ be the oxidation state of

$Cr$ .

$\displaystyle 2(+1) +X + 4(2-) = 0 \\ 2 +X - 8 = 0 \\ X - 6 = 0 \\ X = +6$

The formation of nitrous oxide from nitrogen and oxygen is the example for

0%
Decomposition reaction
0%
Chemical combination of one element and one compound
0%
Chemical combination of two compounds
0%
Chemical combination of two elements

What is the oxidation number of each atom in KH?

0%
+1, -1
0%
-1, +1
0%
-1, -1
0%
+1,+1
0%
Not enough information given.

When tin(IV) chloride is treated with excess of conc. hydrochloric acid, the complex ion

$(SnCl_6)^{2-}$ is formed. The oxidation state of tin in this complex ion is?

0%
$+4$
0%
zero
0%
$-2$
0%
$-4$

Explanation

Let oxidation state of

$Sn$ is

$x$ and we know oxidation of

$Cl$ is

$-1$ , so

$x+6(-1)=-2, x=+4$ .

The oxidation states exhibited by hydrogen in its various molecular forms/compounds are :

0%
-1 only
0%
zero only
0%
+1,-1 and 0
0%
+1 only

Explanation

Oxidation state of hydrogen in

$H_2$ is

$0$

Oxidation state of hydrogen in

$HCl$ is

$+1$

Oxidation state of hydrogen in

$LiH$ is

$-1$

Therefore, hydrogen can exhibit

$-1,0,+1$ oxidation states in various compounds.

"Nitric oxide reacts with oxygen to form reddish brown nitrogen dioxide gas". Choose the correct equation for the above reaction

0%
$2NO + O_{2} \rightarrow 2NO_{2}$
0%
$N_{2}O + O_{2} \rightarrow NO_{2} + NO$
0%
$NaNO_{3} \xrightarrow {heat} 2NaNO_{2} + O_{2}$
0%
$2NO_{3} + O_{2} \rightarrow 2NO_{2} + 2O_{2}$

Explanation

$NO$ : Nitric oxide

$NO_2$ : Nitrogen dioxide gas; It has a reddish brown colour.

Therefore, equation is

$2NO+O_2\longrightarrow 2NO_2$ .

"Reaction of potassium iodide solution with lead nitrate solution" is the example of :

0%
Chemical replacement reaction
0%
Neutralisation reaction
0%
Precipitation reaction
0%
Chemical double decomposition reaction

The reaction,

$AB\rightarrow A + B$ is a:

0%
Decombination reaction
0%
Combination reaction
0%
Chemical composition
0%
Decomposition reaction

Explanation

A decomposition reaction is a type of chemical reaction in which a single compound breaks down into two or more elements or new compounds. Hence it is an example of decomposition/recombination reaction.

In " $AB \rightarrow A+B$ " the compound $AB$ decompose in $A$ and $B.$

What are the oxidation numbers of the atoms in the compound

$CF_4$ ?

0%
+4, -1
0%
+1, -4
0%
+3, -2
0%
+2, -3
0%
+4, -4

Explanation

The oxidation numbers of the atoms in the compound

$CF_4$ are +4 and -1 for C and F respectively.
Since F is most electronegative atom, it has oxidation number of -1.
Four F atoms corresponds to total oxidation number of

$\displaystyle 4 \times -1 = -4$
This is balanced by +4 oxidation number of C.

Heat is released in :

0%
Exothermic reactions
0%
Endothermic reactions
0%
Both A and B
0%
None of these

The oxidation state of nitrogen is most positive in which of the following compounds?

0%
${NO}_{3}^{-}$
0%
${NO}_{2}^{-}$
0%
${N}_{2}O$
0%
${N}_{2}$
0%
${NH}_{3}$

Explanation

${NO_3}^-:x+(3\times -2)=-1$

$\Rightarrow x-6=-1$

$\Rightarrow x=-1+6$

$\Rightarrow x=+5$ .

${NO_2}^-:x+(2\times-2)=-1$

$\Rightarrow x=-1+4$

$\Rightarrow x=+3$

$N_2O:2x+(-2)=0$

$\Rightarrow 2x=2$

$\Rightarrow x=+1$

$N_2$ : elemental state, so oxidation state is zero.

$NH_3: x+3=0$

$\Rightarrow x=-3$

The oxidation number of the nitrogen atom in the ammonium ion is :

0%
+3
0%
+2
0%
+1
0%
-2
0%
-3

Explanation

Let the oxidation number of nitrogen atom in ammonium ion be

$x$ .

Molecular formula of ammonium ion-

${N{H}_{4}}^{+}$

The oxidation number of hydrogen in ammonium ion is

$+1$ .

As we know that the sum of oxidation state of all the atoms in an ion is equal to the overall charge on that ion.

$\therefore$ For

${N{H}_{4}}^{+}$ -

$x + \left( 4 \times 1 \right) = \left( + \right)1$

$\Rightarrow x + 4 = 1$

$\Rightarrow x = 1 - 4 = -3$

Hence the oxidation number of nitrogen atom in ammonium ion is

$-3$ .

Hence the required answer is

$(E) -3$ .

Alumino thermite process involves reduction of chromium with oxide and aluminium.

0%
True
0%
False

Explanation

$Fe_2O_3 + 2Al \rightarrow Al_2O_3 + 2 Fe$

In Alumino thermite process the aluminium act as reducing agent. In this preocess the oxides of various metals (like chromium, iron, etc.) are converted into their corresponding metals. For example: the reduction of ferrous oxide in presence of aluminium results in the formation of ferrous metal. It is exothermic reaction.

In which of the following does sulphur has the lowest oxidation state?

0%
$H_2SO_4$
0%
$SO_2$
0%
$H_3SO_3$
0%
$H_2S$

Explanation

Sulphur has the lowest oxidation state in

${ H }_{ 2 }S$ that is

$-2$ .

The range of oxidation states exhibited by most elements in group V of the periodic table (N.P. As Sb. Bi) is

0%
-3 to +1
0%
-3 to +3
0%
-3 to +5
0%
-2 to +3
0%
+1 to +5

Explanation

Common oxidation state of group

$5$ elements are

$-3,+3$ and

$+5$ .

Nitrogen shows

$+1,+2,+4$ oxidation state. All oxidation state from

$+1$ to

$+4$ disproportionate in acid solution in case of nitrogen. Phosphorous shows

$+1$ and

$+4$ oxidation states in some oxoacids. All intermediate oxidation state disproportionate in acid and alkali into

$-3$ and

$+5$ . However,

$+3$ oxidation state for

$Ar,Sb,Bi$ become increasingly stable with respect to disproportionation.

In which of the following reactions, the oxidation state of nitrogen is not changed?

0%
Action of very dilute $HN{ O }_{ 3 }$ on magnesium
0%
Action of $conc.HN{ O }_{ 3 }$ on zinc
0%
Action of $conc.HN{ O }_{ 3 }$ on copper
0%
Action of very dilute $HN{ O }_{ 3 }$ on zinc

Explanation

Copper is oxidized by concentrated nitric acid, $HN{ O }_{ 3 }$ , to produce ${ Cu }^{ 2+ }$ ions; the nitric acid is reduced to nitrogen dioxide, a poisonous brown gas with an irritating odor.

$Cu+4HN{ O }_{ 3 }\longrightarrow Cu{ (N{ O }_{ 3 }) }_{ 2 }+2N{ O }_{ 2 }+2{ H }_{ 2 }O$

Hydrogen sulphide gets reduced to sulphur.

0%
True
0%
False

The oxidation number for sulfur is highest in which of the following ions or compounds?

0%
$SO_2$
0%
$SO_3^{2-}$
0%
$SO_4^{2-}$
0%
$S_2O_3^{2-}$
0%
$HSO_3^-$

Explanation

$SO_4^{2-}:x+(-2\times 4)=-2$

$\Rightarrow x-8=-2$

$\Rightarrow x=-2+8$

$\Rightarrow x=6$ .

What is the oxidation state of

$Mn$ in

$KMnO_4$ ?

0%
-7
0%
-3
0%
0
0%
+3
0%
+7

Explanation

Option

$E$ is the correct answer.

${KMnO}_{4}$ dissociates as

${K}^{+}$ and

${MnO}_{4}^{-}$ . Thus we will now calculate the oxidation number of

${Mn}$

Let us suppose it has an oxidation state of X. Thus the equation is-

${+1}$

${+}$

${X}$

${-}$

${8}$

${=}$

${0}$

Thus the value of X is

${+7}$

The equation has been written in that manner because

${KMnO}_{4}$ has an overall oxidation number of 0.

${K}$ has an oxidation number of

${+1}$ and

${O}$ has an oxidation number of

${-2}$ . Thus four oxygen atoms will have an oxidation number of

${-8}$

Oxidation numbers on both the sides should be balanced and hence we get the value to be +7.

In the electrochemical cell shown above, which of the following half-reactions occurs at the anode?

0%
$\displaystyle { Cu }^{ 2+ }+e\rightarrow { Cu }^{ - }$
0%
$\displaystyle Zn\left( s \right) \rightarrow { Zn }^{ 2+ }+2e$
0%
$\displaystyle { Zn }^{ 2+ }+2e\rightarrow Zn\left( s \right)$
0%
$\displaystyle Cu\left( s \right) \rightarrow { Cu }^{ 2+ }+2e$
0%
$\displaystyle { Cu }^{ 2+ }+2e\rightarrow Cu\left( s \right)$
$\displaystyle$

Explanation

At anode, oxidation will take place and

$Zn(s)$ will get oxidised to

$Zn^{2+}$ leaving negatively charged electrons, making anode negatively charged, repelling negatively charged

$NO_3^-$ .

Oxidation number of Cl in

$CaOCl_2$ is ____________.

0%
$-1$ and $+1$
0%
$+2$
0%
$-2$
0%
none of these

Explanation

$Ca(OCl)Cl$ , two

$Cl$ atoms are in different oxidation state i.e., one

$Cl^{-}$ in -1 oxidation state and other as

$OCl^{-}$ in +1 oxidation state.

Which of these half-reactions represents an oxidation?
I.

$Sn^{4+} \rightarrow Sn^{2+}$
II.

$MnO_4^-\rightarrow Mn^{2+}$
III.

$Cl_2\rightarrow ClO_3^-$

0%
I only
0%
II only
0%
III only
0%
I and II only
0%
II and III only

Explanation

Oxidation state of

$Sn$ changes from

$+4$ to

$+2$

Oxidation state of

$Mn$ changes from

$+7$ to

$+2$

Oxidation state of

$Cl$ changes from

$0$ to

$+7$

Therefore reaction

$III$ represents oxidation.

Which of these types of chemical reactions is NOT a redox reaction?

0%
Decomposition into elements
0%
Single replacement
0%
Double replacement
0%
Combustion
0%
Synthesis of two eiements

Statement I : The oxidation state of Cr in

$Al_2(Cr_2O_7)_3$ is +3.
Statement II : as a neutral compound, the sum of oxidation numbers of all the atoms must equal zero.

0%
Statement 1 and Statement 2 are correct and Statement 2 is the correct explanation of Statement 1
0%
Both the Statement 1 and Statement 2 are correct and Statement 2 is NOT the correct explanation of Statement 1
0%
Statement 1 is correct but Statement 2 is not correct
0%
Statement 1 is not correct but Statement 2 is correct
0%
Both the Statement 1 and Statement 2 are not correct.

Explanation

Oxidation state of

$Al$ is

$+3$ . And oxidation number of

$O$ is

$-2$ . Thus, oxidation state of

$Cr$ can be calculated by considering it as

$X$ . Thus-

(

$2$

$*$

$3$ )

$+$

$6X$

$+$ [

$21$

$*$ (

$-2$ )]

$=$

$0$

The value of

$X$ obtained is

$+6$ . Oxidation state of

$Cr$ in the above mentioned molecule is

$+6$ .

In a neutral compound, the sum of oxidation numbers of all the atoms must be equal to zero. If it's not zero, it can't be neutral.

Thus, statement 1 is not correct. Statement 2 is correct.

Option

$D$ is the correct answer.

What is the oxidation number of

$Mn$ in

$Mn{O}_{2}$ ?

0%
$+2$
0%
$+4$
0%
$-2$
0%
$-4$

Explanation

The oxidation number of simple ions is equal to the charge on the ion. The oxidation number of

$Mn$ in

$MnO_2$ is

$Mn^+4$ .

Hence option

$B$ .

What is the oxidation number of lead in

$Pb{F}_{2}$ ?

0%
$0$
0%
$+2$
0%
$+1$
0%
$-1$

Explanation

$PbF_2$

The total charge present is

$0$ .

Oxidation no. of Fluorine is

$-1$ .

Oxidation no. of lead be

$x$ .

So,

$x+2(-1)=0\\x-2=0\\x=2$

So, oxidation no. of lead in

$PbF_2$ is

$+2$ .

What is the oxidation state of bromine in

$Al{Br}_{3}$ ?

0%
$-1$
0%
$+3$
0%
$0$
0%
$-3$

Explanation

Oxidation state of

$Br$ in

$AlBr_3: +3+3x=0$

$\Rightarrow 3x=-3$

$\Rightarrow x=-1$ .

What is the oxidation number for As in the chemical formula for arsenic acid,

$H_3AsO_4$ ?

0%
+1
0%
+2
0%
+3
0%
+4
0%
+5

Explanation

Oxidation number of

$As$ in

$H_3AsO_4$ :

$\Rightarrow 3+x+(4\times -2)=0$

$\Rightarrow 3+x-8=0$

$\Rightarrow x=8-3$

$\Rightarrow x=+5$ .

In which of the following compounds does bromine have the highest positive oxidation state?

0%
$\displaystyle HBr$
0%
$\displaystyle Br{ F }_{ 3 }$
0%
$\displaystyle Br{ O }_{ 2 }$
0%
$\displaystyle NaBrO$
0%
$\displaystyle NaBr{ O }_{ 3 }$

Explanation

Compound Oxidation state of Bromine

$HBr$

$-1$

$BrF_3$

$+3$ [

$\because$ Fluorine is highest electronegative element]

$BrO_2$

$+4$ [

$\because$ Oxygen electronegative value is more than electronegative value of

$Br$ ]

$NaBrO$

$+1$

$NaBrO_3$

$+5$

In which substance does manganese have the highest oxidation number?

0%
$LiMnO_{4}$
0%
$K_{2}MnO_{4}$
0%
$MnO_{2}$
0%
$Mn(NO_{3})_{2}$

Explanation

$LiMnO_4\\+1+x+4(-2)=0,\\x-7=0,\\x=7.$

$K_2MnO_4\\2(1)+x+4(-2)=0,\\x-6=0,\\x=6.$

$MnO_2\\x+2(-2)=0,\\x+4.$

$Mn{(NO_3)}_2\\x+2(-1)=0,\\x=2.\\$

Out of

$4$ , Manganese in

$LiMnO_4$ has the highest oxidation number.

Which of the following are redox reactions?
I.

$PCl_{3} + Cl_{2}\rightarrow PCl_{5}$
II.

$Cu + 2AgNO_{3} \rightarrow Cu(NO_{3})_{2} + 2Ag$
III.

$CO_{2} + 2LiOH \rightarrow Li_{2}CO_{3} + H_{2}O$
IV.

$FeCl_{2} + 2NaOH \rightarrow Fe(OH)_{2} + 2NaCl$

0%
II
0%
I and II
0%
I, II, and III
0%
I, II, III and IV

Explanation

If in a reaction both, oxidation and reduction occurs, then it is called redox reaction.

$Cu+2AgNO_3\longrightarrow Cu{(NO_3)}_2+2Ag\\{\text{ }} \downarrow \quad \quad\downarrow\quad\quad\quad\quad\quad\quad \downarrow \quad\quad\quad\quad\downarrow$

$0$

$x-1=0\\x=1$

$x+2(-1)=0\\x=+2$

$0$

Oxidation state of

$Cu$ changes from

$0$ to

$+$ , it is oxidation.

$Ag$ reduced from

$+1$ to

$0$ , it is reduction.

So, all total it is a redox reaction.

The oxidation number of chromium in

$K_{2}CrO_{4}$ is?

0%
$2+$
0%
$3+$
0%
$6+$
0%
$8+$

Explanation

$K_2CrO_4$ ,

Oxidation no. of

$K$ is

$+1.$

Oxidation no. of

$O$ is

$-2.$

Oxidation no. of

$Cr$ be

$x$ .

$2(1)+x+4(-2)=0,\\ 2+x-8=0,\\ x-6=0,\\ x=6.$

So, Oxidation no. of

$Chromium$ in

$K_2Cr_2O_4$ is

$+6$ .

What is one of the spectator ions in the following reaction?

$Pb(NO_{3})_{2} (aq) + 2KI(aq) \rightarrow PbI_{2} (s) + 2KNO_{3}(aq)$

0%
$NO_{3}{^-}$
0%
$I^{-}$
0%
$PbI_{2}$
0%
$Pb^{+2}$

Explanation

A precipitation reaction occurs when two ionic compounds react to produce a precipitate and an insoluble substance which stays in solution as spectator ion.

$\longrightarrow$ Ions that are not involved in the actual chemical reaction are called spectator ions.

$\longrightarrow$ Net ionic equation of given reaction is

$Pb^{+2}+2I^-\longrightarrow PbI_2$ It shows only the ions which involves in the chemical reaction,

Spectator ions:-

$K^+$ and

${NO_3}^-$ .

Which of the following reactions represents an oxidation-reduction reaction?

0%
$2NaCl_{(aq)} + K_{2}CO_{3(aq)} \rightarrow Na_{2}CO_{3(aq)} + 2KCl_{(aq)}$
0%
$CaCO_{3(aq)} \xrightarrow {\Delta} CaO_{(s)} + CO_{2(g)}$
0%
$2Mg_{(s} + O_{2(g)} \rightarrow 2MgO_{(s)}$
0%
$3FeCl_{2(aq)} + 2Cu_{3}PO_{4(aq)} \rightarrow Fe_{3}(PO_{4})_{2(aq)} + 6CuCl_{(aq)}$

Explanation

In a single reaction

$Sodium(I)$ in

$NaCl$ was changed to

$Sodium(II)$ in

$Na_2CO_3$ means it is oxidised and

$Potassium(II)$ in

$K_2CO_3$ is reduced to

$K(I)$ in

$KCl$ . It is reduced. These two occur in a single reaction. So, it represents oxidation-reduction reaction.

Your group was asked to identify which types of reactions can be represented as redox reactions. Everyone has given an examples of what they think is correct.
Check their answer and decide who is right.

0%
Crystal Lynn says synthesis reactions can be classified as redox: $Na_{(s)} + Cl_{2(g)} \rightarrow 2NaCl_{(s)}$
0%
Barry Um says decomposition reactions can be classified as redox: $H_{2}O_{(l)}\rightarrow H_{2(g)} + O_{2(g)}$
0%
Barry's brother, Cal C. Um states that single replacement reactions can be classified as redox: $Ag_{(s)} + NaCl_{(aq)} \rightarrow Na_{(s)} + AgCl_{(aq)}$
0%
Molly Cule state everyone in the group is correct. Redox reactions can be classified as each of the examples cited.

Explanation

$Crystal$

$Lynn$ is correct because,

$2Na_{(s)}+{Cl_2}_{(g)}\longrightarrow 2NaCl_{(s)}$ is a redox reaction.

Oxidation state of

$Na$ and

$Cl$ in reactant side is is

$0$ .

Because they are in free state.

But in right side,

$Na$ is in

$+1$ state and

$Cl$ is in

$-1$ state.

$Na$ changed from

$0$ to

$+1\Longrightarrow Oxidised$

$Cl$ changed from

$0$ to

$-1\Longrightarrow Reduced$

Oxidation and reduction, both occured in a reaction. So, it is a redox reaction.

Select the non-redox reaction with the correct justification.

0%
$2Na + Cl_{2}\rightarrow 2NaCl$ ; electrons are not transferred.
0%
$Fe + Cu^{2+} \rightarrow Fe^{2+} + Cu$ ; electron are transferred.
0%
$Ag^{+} + Cl^{-}\rightarrow AgCl$ ; electron are not transferred.
0%
$CH_{4} + 2O_{2} \rightarrow CO_{2} + 2H_{2}O$ ; electron are transferred.

Explanation

In a redox reaction both of the chemical species must get oxidized or reduced and no

$e^-$ donated or gained.

$A)$

$2N a^0+Cl^0_2\longrightarrow 2N^{+1}Cl^{-1}$

$Na$ getting oxidized from

$0$ (in

$Na$ ) to

$+1$ (in

$NaCl$ ).

$Cl$ is getting reduced from

$0$ (in

$Cl_2$ ) to

$-1$ (in

$NaCl$ )

$\therefore$ It is a redox reaction.

$B)$

$Fe+Cu^{2+}\longrightarrow Fe^{2+}+Cu$

$Fe:$ Oxidized

$Cu:$ reduced.

$\therefore$ This is a redox reaction.

$C)$

$Ag^++Cl^-\longrightarrow A^{+1}gCl^{-1}$

Neither

$Ag^+$ nor

$Cl^-$ is getting oxidized or reduced.

$\therefore$ It is not a redox reaction.

$D)$

$C^{+4}H^{+1}_4+2O_2\longrightarrow C^{+4}O_2+2H^{+1}_2O$

Carbon is getting oxidized from

$-4$ (in

$CH_4$ ) to

$+4$ (in

$CO_2$ ), but no other species is getting reduced. Therefore not a redox reaction.

$I$ . The oxidation state of carbon in the compounds

${CH}_{4},{C}_{2}{H}_{6}$ and

${C}_{2}{H}_{2}$ is

$+4$ .

$II$ . Carbon has four valence electrons.

0%
Statement $I$ is true, Statement $II$ is true
0%
Statement $I$ is true, Statement $II$ is false
0%
Statement $I$ is false, Statement $II$ is true
0%
Statement $I$ is false, Statement $II$ is false

Explanation

$I.$ For covalent compounds, there is no oxidation state.

$II.$ Carbon has

$4$ valence electrons. see the figure.

1175032_526237_ans_153c16710f854f9c8c7e4a5083a6d636.png

What is the oxidation number of

$Br$ in the compound

$RbBr{O}_{4}$ ?

0%
$-1$
0%
$+7$
0%
$+1$
0%
$+4$

Explanation

$RbBrO_4$

Total charge

$=0.$

$Rb$ is a

$group$

$I$ element. So, its oxidation no.

$=+1$ .

Oxidation no. of oxygen

$=-2$ .

Let, Oxidation no. of

$Br$ be

$x$ .

So,

$1+x+4(-2)=0\\1+x-8=0\\x-7=0\\x=7$

So, oxidation state of

$Bromine$ is

$+7$ .

Balanced chemical equation imply which of the following?

0%
Numbers of atoms are conserved in a chemical change
0%
Numbers of molecules are conserved in a chemical change
0%
Volume is conserved in a chemical change
0%
'A' and 'B'
0%
'B' and 'C'

In which of the following compounds does nitrogen have the highest oxidation number?

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Nitrogen Gas
0%
Aluminum Nitride
0%
Potassium Nitrate
0%
Nitrous Acid

Explanation

In Nitrogen gas, oxidation no. of Nitrogen is

$0$ . Because , it exist in free state.

In Aluminium Nitrate, it is

$-3\Longrightarrow AlN\Longrightarrow +3+x=0=-3.$

In Potassium Nitrate,

$KNO_3\Longrightarrow 1+x+3(-2)=0=+5$ .

$HNO_2$

$\Longrightarrow 1+x+2(-2)=0,\\x-3=0,\\x=3.$

So, highest oxidation no. is in Potassium nitrate.

Which expression gives the correct value for the standard potential for a gold-rhodium voltaic cell?

Half-reaction	$E^{\circ}(V)$
$Rh_{(aq)}^{+3} + 3e^{-} \rightarrow Rh_{(s)}$	$0.76\ V$
$Au_{(aq)}^{+} + e^{-} \rightarrow Au_{(s)}$	$1.69\ V$

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$1.69\ V + 0.76\ V$
0%
$1.69\ V - 0.76\ V$
0%
$3(1.69)\ V + 0.76\ V$
0%
$3(1.69)\ V - 0.76\ V$

Explanation

We know that, metal having higher reduction potential undergoes reduction;

so, Au will undergo reduction and Rh will undergo oxidation.

$E^0_{cell}$ = 1.69 V - 0.76 V

Find the oxidation number of

$N$ &

$Cl$ in

$NOCl_4$ respectively is:

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$+3, +6$
0%
$+2, +6$
0%
$+3, +7$
0%
$+2, +7$

Explanation

$NOCl_4$ is made up of

$NO^+$ and

$ClO_4^-$

$NO^+$ x+1(-2)=+1
x=+3

$ClO_4^-$ x-8=-1
x=+7

Which of the following represents the oxidation half-reaction for the unbalanced redox equation?

$Cu + HNO_{3} \rightarrow Cu (NO_{3})_{2} + NO_{2}$

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$Cu^{2+} + 2e^{-} \rightarrow Cu$
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$Cu \rightarrow Cu^{2+} + 2e^{-}$
0%
$NO_{3}^{-} \rightarrow NO_{2} + 1e^{-}$
0%
$2H^{+} + NO_{2}^{-} + 1e^{-} \rightarrow NO_{2} + H_{2}O$

Explanation

$Cu+HNO_3 \longrightarrow Cu(NO_3)_2+NO_2$

As Oxidation of

$Cu$ is taking place in the reaction.

Thus the oxidation half-cell reaction will be

$Cu \longrightarrow Cu^{+2}+2e^-$

The given figure represents a

$Zn-Cu$ electrochemical cell:
Based on the figure, which of the following represents the balanced chemical reaction?

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$Zn_{(aq)}^{2+} + Cu_{(s)} \rightarrow Zn_{(s)} + Cu_{(aq)}^{2+}$
0%
$Zn_{(s)} + Cu_{(aq)}^{2+} \rightarrow Zn_{(aq)}^{2+} + Cu_{(s)}$
0%
$Zn_{(s)} + Cu_{(s)} \rightarrow Zn_{(aq)}^{2+} + Cu_{(aq)}^{2+}$
0%
$Zn_{(aq)}^{2+} + Cu_{(aq)}^{2+} \rightarrow Zn_{(s)} + Cu_{(s)}$

Explanation

$Zn_{(s)}+Cu^{+2}_{(aq.)} \longrightarrow Zn^{+2}_{(aq.)}+Cu_{(s)}$

$\longrightarrow$ This is also called Galvanic cell reaction.

$\longrightarrow$ In this

$Cu^{+2}$ converts to

$Cu$ due to high standard reduction potentials than

$Zn$ .

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 5 - MCQExams.com

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Practice Class 11 Medical Chemistry Quiz Questions and Answers

CBSE Questions for Class 11 Medical Chemistry Redox Reactions Quiz 5 - MCQExams.com

0:0:1

Practice Class 11 Medical Chemistry Quiz Questions and Answers

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