Chemistry 2 Marks Questions

Then, we have

2(+1) + 2(x) + 7(-2) = 0

⇒ 2 + 2x - 14 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is +6.

Then, we have

4(+1) + 2(x) + 7(-2) = 0

⇒ 4 + 2x - 14 = 0

⇒ 2x = +10

⇒ x = +5

Hence, the oxidation number of P is +5.

Then, we have

1(+1) + 1(x) + 4(-1) = 0

⇒ 1 + x - 4 = 0

⇒ x = +3

Hence, the oxidation number of B is +3.

Then, we have

1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2)=0

⇒ 1 + 3 + 2x - 16 + 24 - 24 = 0

⇒ 2x = 12

⇒ x = +6

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

1(+1) + 1(+3) + 2(x) + 8(-2) = 0

⇒ 1 + 3 + 2x - 16 = 0

⇒ 2x = 12

⇒ x = +6

Hence, the oxidation number of S is + 6.

Then, we have

2(+1) + x + 4(-2) = 0

⇒ 2 + x - 8 = 0

⇒ x = +6

Hence, the oxidation number of Mn is +6.

Then, we have

1(+1) + 1(+1) + 1(x) + 4(-2) = 0

⇒ 1 + 1 + x - 8 = 0

⇒ x = +6

Hence, the oxidation number of S is +6.

Then, we have

(+2) + 2(x) = 0

⇒ 2 + 2x = 0

⇒ x = -1

Hence, the oxidation number of O is -1.

Hence, the oxidation number of P is +5.

1. Let the oxidation number of phosphorus is x.

$\stackrel{\text{x}}{\text{HPO}}^{2-}_3$

+1 + x + (-2) × 3 = -2

+1 + x - 6 = -2

x - 2 = -2

x = -2 + 5

x = +3

Thus, O.S. of phosphorous is +3.

2. $\stackrel{\text{x}}{\text{P}}\text{O}^{3-}_4$

x + (-2) × 4 = -3

x - 8 = -3

x = -3 + 8

x = +5

Thus, O.S. of phosphorous in this ion is +5.

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)}+\text{Fe}^{2+}\text{(aq)}\xrightarrow{ \ \ \\ \ \ \\}\text{Cr}^{3+}\text{(aq)}+\text{Fe}^{3+}\text{(aq)}\\ +6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +3\ \ \ \ \ \ \ \ \ \ \ \ \ \ +3$

Step 2: The oxidation state of Cr decreases by 3 per chromium atom, total decrease is 6 for two chromium atoms, oxidation state of Fe changes from +2 to +3, i.e., increases by 1, therefore, to equalize the increase and decrease, we multiply Fe²⁺ by 6 and Cr₂O₇^2- by 1.

Step 3: Balancing Cr' and Fe on both sides.

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)}+6\text{Fe}^{2+}\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ }2\text{Cr}^{3+}\text{(aq)}+6\text{Fe}^{3+}\text{(aq)}$

Step 4: To balance oxygen, we add 7 molecules of H₂O on RHS.

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)}+6\text{Fe}^{2+}\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ }2\text{Cr}^{3+}\text{(aq)}+6\text{Fe}^{3+}\text{(aq)}+7\text{H}_2\text{O}\text{(l)}$

Step 5: To balance hydrogen, we add 14H⁺ on LHS and we get balanced equation.

$\text{Cr}_2\text{O}_7^{2-}\text{(aq)}+6\text{Fe}^{2+}\text{(aq)}+14\text{H}^+\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ }2\text{Cr}^{3+}\text{(aq)}+6\text{Fe}^{3+}\text{(aq)}+7\text{H}_2\text{O}\text{(l)}$

1. It is because 'Cu' is less reactive than Fe, so cannot displace Fe from FeSO₄ solution.

1. Zinc is more reactive than Cu but Ag is less reactive than Cu.
2. because Cu is more reactive than Ag.₂)₃ It is due to formation of Cu (NO₃)₂ because Cu is more reactive than Ag.$\text{Cu}\text{(s)}+\text{AgNO}_3\text{(aq)}\xrightarrow{ \ \ \ \ \ \ }\text{Cu(NO}_3)_2\text{(aq)}+2\text{Ag}\text{(s)}$

$\text{MnO}_4^-\text{(aq)}+\text{Br}\text{(aq)}\xrightarrow{ \ \ \ \ \ }\text{MnO}_2\text{(aq)}+\text{BrO}_3^-\text{(aq)}\\ \ +7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ +5$

Step 2: $\text{MnO}_4^-$ is oxidant because its oxidation state is decreasing. Br^- is reductant because its oxidation state is increasing.

Step 3: Oxidation state of Mn is decreasing by 3. Oxidation state of Br is increasing by 6. To equalize increase and decrease, multiply $\text{MnO}_4^-$ by 2 and Br^- by 1 we get.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{Br}_2^-\text{(aq)}$

Step 4: Now for balncaing oxygen, we add 1 molecule of H₂O on RHS.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}$

Step 5: As the reaction is taking place in basic medium to balance hydrogen, add 2H₂O molecules on LHS and 2OH^- on RHS.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}+2\text{OH}^-\text{(aq)}$

It can be seen 1 molecule of H₂O gets cancelled on both sides, we get.

$2\text{MnO}_4^-\text{(aq)}+\text{Br}^-\text{(aq)}\\\xrightarrow{ \ \ \ \ \ \ \ }\text{MnO}_2\text{(s)}+\text{BrO}_3^-\text{(aq)}+\text{H}_2\text{O}\text{(l)}+2\text{OH}^-\text{(aq)}$

is a balanced equation.

Multiply P by l, HNO₃ by 5, we get$\text{P}+5\text{HNO}_3\xrightarrow{ \ \ \ \ \ \ \ \ }\text{H}_3\text{PO}_4+5\text{NO}_2+\text{H}_2\text{O}$

$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$

The reaction taking place at Zn electrode can be represented as:

$\text{Zn}_{(\text{s})}\rightarrow\text{Zn}^{2+}_{(\text{aq})}+2\text{e}^-$

And the reaction taking place at Ag electrode can be represented as:

$\text{Ag}^+_{(\text{aq})}+\text{e}^-\rightarrow\text{Ag}_{(\text{s})}.$

1. At Cathode: $2\text{H}^+\text{(aq)}+2\text{e}^-\xrightarrow{ \ \ \ \ \ \ \ }\text{H}_2\text{(g)}$

2. At Anode: $\text{H}_2\text{(g)}\xrightarrow{ \ \ \ \ \ }2\text{H}^+\text{(aq)}+2\text{e}^-$

1. $\stackrel{{\text{x}-2}}{\hbox{ClO}_2}$

$\text{x}-4=0$

$\Rightarrow\text{x}=+4$

2. $\stackrel{{\text{x}-2}}{\hbox{ClO}_3^-}$

$\text{x}-2\times3=-1$

$\text{x}-6=-1$

$\Rightarrow\text{x}=+5$

$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$

Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and the leaving electrons accumulate on this electrode.

$\text{Zn}|\text{Zn}^{2+}_{(\text{aq})}||\text{Ag}^{+}_{(\text{aq})}|\text{Ag}$

Ions are the carriers of current in the cell.

3 + 4 + 3x - 14 = 0

3x = 7; $\text{x} = \frac{7}{3}$

The oxidation number of each element in the given reaction can be represented as:

$\stackrel{{+3\ -1}}{4\ \ \ \ \ \hbox{BCl}_{3(\text{g})}}+\stackrel{{+1\ \ +3\ -1}}{3\ \ \ \ \hbox{LiAlH}_{4(\text{s})}}\ \rightarrow\stackrel{{-3}}{2\ \ \hbox{B}_2}\stackrel{{+1}}{\ \ \ \ \ \hbox{H}_{6(\text{g})}}+\stackrel{{+1}}{3\hbox{Li}}\stackrel{{-1}}{\ \ \ \ \hbox{Cl}_{(\text{s})}}+\stackrel{{+3}}{3\hbox{Al}}\stackrel{{-1}}{\ \ \ \ \ \ \hbox{Cl}_{3(\text{s})}}$

In this reaction, the oxidation number of B decreases from +3 in BCl₃ to –3 in B₂H₆. i.e., BCl₃ is reduced to B₂H₆. Also, the oxidation number of H increases from –1 in LiAlH₄ to +1 in B₂H₆ i.e., LiAlH₄ is oxidized to B₂H₆. Hence, the given reaction is a redox reaction.

Let us write the oxidation number of each element in the given reaction as:

$\stackrel{{+3}}{\ \ \ \ \hbox{Fe}_2}\stackrel{{-2}}{\ \ \ \ \ \hbox{O}_{3(\text{s})}}+\stackrel{{+2-2}}{3\ \ \ \hbox{CO}_{(\text{g})}}\ \rightarrow\stackrel{{0}}{\ \ \ \ \hbox{2Fe}_{(\text{s})}}+\stackrel{{+4-2}}{3\ \ \ \ \ \hbox{CO}_{2(\text{g})}}$

Here, the oxidation number of Fe decreases from +3 in Fe₂O₃ to 0 in Fe i.e., Fe₂O₃ is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in CO₂ i.e., CO is oxidized to CO₂. Hence, the given reaction is a redox reaction.

The oxidation number of each element in the given reaction can be represented as:

Let us write the oxidation number of each element involved in the given reaction as:

1. Salt bridge.

1. Completes internal circuit.
2. It maintains electroneutrality.

1.  $\stackrel{{+4}}{\hbox{C}\text{a}}\stackrel{{-2+2}}{\hbox{CO}_3}(\text{s})\xrightarrow{\ \ \ \ \ \ \ \ }\stackrel{{+2}}{\hbox{Ca}}\stackrel{{-2}}{\hbox{O}}(\text{s})+\stackrel{{+4-2}}{\hbox{CO}_2}\text{(g)}$

It is not a redox reaction.

2. $\stackrel{{+2+6}}{\hbox{Fe}}\stackrel{{-2}}{\hbox{SO}_4}\xrightarrow{\ \ \ \Delta \ \ \ \ \ }\stackrel{+3}{\hbox{Fe}}\stackrel{-2}{\hbox{O}_3}+\stackrel{{_-2}}{\hbox{SO}_2}+\stackrel{{+6-2}}{\hbox{SO}_3}$ 

1. Let oxidation number of Fe in Fe₃O₄ be x

$\therefore3\text{x}-8=0$

$\Rightarrow\text{x}=\frac{8}{3}$

It is the average oxidation number of Fe²⁺ and 2Fe³⁺

1. $\text{MnO}_4^-+\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{Mn}^{2+}$ in acidic medium

$[\text{Fe}^{2+}\xrightarrow{ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-]\times5\ \cdots\text{(i)}$

$5\text{e}^-+8\text{H}^++\text{MnO}_4^-\xrightarrow{ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O} \ \cdots\text{(ii)}$

Adding (i) and (ii), we get 

$5\text{Fe}^{2+}+8\text{H}^++\text{MnO}_4^-\xrightarrow{ \ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O} \ +5\text{Fe}^{3+}$

2. Ag < Mg < K is increasing order of reducing power.

1. Zn is reducing agent because it is losing electrons to form Zn²⁺ i.e. oxidation state is increasing from 0 to +2. $\text{NO}_3^-$ is oxidising agent because oxidation state of N is decreasing from +5 to -3, i.e. it is gaining electrons.
2. I­ is oxidising agent because it is gaining electrons. Its oxidation state is decreasing from 0 to -1 whereas H₂S is reducing agent, the oxidation state of 'S' is increasing from -2 to 0 by losing electrons.

1. It maintains electroneutrality.
2. It completes internal circuit.

1. Redox couple. It is defined as having together the oxidised and reduced forms of a substance taking part in an oxidation or reduction half reaction e.g. $\frac{\text{Zn}^{2+}}{\text{Zn}}$ and $\frac{\text{Cu}^{2+}}{\text{Cu}}$ form a redoxc couple. application. These are used to make electrochemical cell.
2. $2\text{K}\xrightarrow{ \ \ \ \ }2\text{K}^++2\text{e}^-$ (oxidation half reaction)

$\text{Cl}_2+2\text{e}^-\xrightarrow{ \ \ \ \ }2\text{Cl}^-$ (reduction half reaction)

1. HNO₃

+1 + x - 6 = 0

x = +5

2. NO

x - 2 = 0

x = +2

+1 -1

3. NH₄Cl

x + 4 -1 = 0

x = -3

4. N₂

N₂ has oxidation state 

HNO₃ > NO > N₂ > NH₄Cl

$\Rightarrow\text{x}=+5$