Redox Reactions

To determine the oxidation states of nitrogen in each compound, we consider the overall charge of the compound and the known oxidation states of other elements involved, typically oxygen.

NO (Nitric Oxide):

Oxygen generally has an oxidation state of $-2$.

Let the oxidation state of nitrogen in NO be $x$.

The equation: $x + (-2) = 0$.

Solving for $x$, we find $x = +2$.

NO$_2$ (Nitrogen Dioxide):

Oxygen has an oxidation state of $-2$.

Let the oxidation state of nitrogen be $x$.

The equation: $x + 2(-2) = 0$.

Solving for $x$, we find $x = +4$.

N$_2$O (Dinitrogen Monoxide or Nitrous Oxide):

Oxygen has an oxidation state of $-2$.

Let each nitrogen atom have an oxidation state of $x$.

The equation: $2x + (-2) = 0$.

Solving for $x$, we find $x = +1$.

NO$_3^-$ (Nitrate Ion):

Oxygen has an oxidation state of $-2$.

The overall charge of the ion is $-1$.

Let the oxidation state of nitrogen be $x$.

The equation: $x + 3(-2) = -1$.

Solving for $x$, we find $x = +5$.

Therefore, the oxidation states of nitrogen are:

NO: $+2$

NO$_2$: $+4$

N$_2$O: $+1$

NO$_3^-$: $+5$

The order of oxidation states from highest to lowest is:

NO$_3^-$ ($+5$) > NO$_2$ ($+4$) > NO ($+2$) > N$_2$O ($+1$)

Thus, the correct option is:

Option D

NO$_3^-$ > NO$_2$ > NO > N$_2$O

NaH is a strong H^– (hydride) donor. Hence cannot be used as an oxidising agent.


In Pyridine, the lone pair of ‘N’ is localised, makes it basic.

Bunsen Burner and measuring cylinder is not required for titration.

In H₃PO₄, P is present in +5 oxidation state and it can act as reducing agent only.

N₂ + O₂ $ \to $ 2NO

during the reaction, oxidation of nitrogen take place from 0 to 2 and reduction of oxygen take place from 0 to –2. It means this reaction is redox reaction.

3O₂ $ \to $ 2O₃ (Non - redox reaction)

2NaOH + H₂SO₄ $ \to $ Na₂SO₄ + 2H₂O (neutralization reaction)

[Co(H₂O)₆]Cl₃ + 3AgNO₃ $ \to $ 3AgCl $ \downarrow $ + [Co(H₂O)₆](NO₃)₃

Reaction of [CO(H₂O)₆]Cl₃ with AgNO₃ is not redox reaction. It is a precipitation reaction.

Potasisum has an oxidation of +1 (only) in combined state.

In disproportionation reaction one element undergoes both oxidation and reduction.

Here disproportionation reaction is :

In N₂O oxidation states of nitrogen = +1

In NO oxidation states of nitrogen = +2

In N₂O₃ oxidation states of nitrogen = +3

In NO₂ oxidation states of nitrogen = +4

$ \therefore $ Equivalent of KMnO₄ = Eq of FeC₂O₄ + Eq of Fe₂(C₂O₄)₃ + FeSO₄

$ \Rightarrow $ moles of KMnO₄ $ \times $ 5 = 3 $ \times $ 1 + 6 $ \times $ 1 + 1 $ \times $ 1

$ \Rightarrow $ moles of KMnO₄ = ${{10} \over 5}$ = 2

The reaction between sodium hydroxide (NaOH) and oxalic acid (H₂C₂O₄) is as follows :

$ \text{H}_2\text{C}_2\text{O}_4 + 2\text{NaOH} \rightarrow \text{Na}_2\text{C}_2\text{O}_4 + 2\text{H}_2\text{O} $

We can see that 1 mole of oxalic acid ($\text{H}_2\text{C}_2\text{O}_4$) reacts with 2 moles of sodium hydroxide (NaOH).

Given that 50 mL of 0.5 M oxalic acid is needed to neutralize the sodium hydroxide, we can find the number of moles of oxalic acid that reacted :

$ \text{Moles of H}_2\text{C}_2\text{O}_4 = \text{Molarity} \times \text{Volume (L)} = 0.5 \, \text{mol/L} \times 0.05 \, \text{L} = 0.025 \, \text{mol} $

Since 1 mole of $\text{H}_2\text{C}_2\text{O}_4$ reacts with 2 moles of $\text{NaOH}$, the number of moles of $\text{NaOH}$ in 25 mL solution would be twice that of $\text{H}_2\text{C}_2\text{O}_4$ :

$ \text{Moles of NaOH} = 0.025 \, \text{mol} \times 2 = 0.05 \, \text{mol} $

Now, to find the mass of NaOH, we multiply the number of moles by the molar mass of NaOH (40 g/mol) :

$ \text{Mass of NaOH} = \text{Number of moles} \times \text{Molar mass} = 0.05 \, \text{mol} \times 40 \, \text{g/mol} = 2 \, \text{g} $

However, the question asks for the amount of $\text{NaOH}$ in 50 mL of the given solution. Since we've found the amount in 25 mL, we just need to double our result to find the amount in 50 mL :

$ 2 \, \text{g} \times 2 = 4 \, \text{g} $

So, the correct answer is Option B : 4 g.

n_CaSO4 $ \times $ Van't hoff factor = n_CaCO3 $ \times $ Van't hoff factor

$ \Rightarrow $ 10^-3 $ \times $ 2 = n_CaCO3 $ \times $ 2

$ \Rightarrow $ n_CaCO3 = 10^-3 mol in 1 L

$ \therefore $ w_CaCO3 = 100 $ \times $ 10^-3 g CaCO₃ in 1 L solution

$ \therefore $ hardness in terms of CaCO₃

= ${{{w_{CaC{O_3}}}} \over {{w_{Total}}}} \times {10^6}$

= ${{100 \times {{10}^{ - 3}}} \over {1000}} \times {10^6}$

= 100 ppm

The balanced reaction of oxalate with permanganate in acidic medium is :

$5 \, \text{C}_2\text{O}_4^{2-} + 2 \, \text{MnO}_4^- + 16 \, \text{H}^+ \rightarrow 10 \, \text{CO}_2 + 2 \, \text{Mn}^{2+} + 8 \, \text{H}_2\text{O}$

In this redox reaction, the oxalate ion $\text{C}_2\text{O}_4^{2-}$ is oxidized to carbon dioxide $\text{CO}_2$, and the permanganate ion $\text{MnO}_4^-$ is reduced to manganese(II) ion $\text{Mn}^{2+}$.

The oxidation half-reaction, representing the change for the oxalate ion, is :

$\text{C}_2\text{O}_4^{2-} \rightarrow 2 \, \text{CO}_2 + 2 \, e^-$

From this half-reaction, it's clear that each oxalate ion produces two CO₂ molecules and releases two electrons in the process.

So, the number of electrons involved in producing one molecule of CO₂ is :

$\frac{2 \, e^-}{2 \, \text{CO}_2} = 1 \, e^-/\text{CO}_2$

Therefore, the correct answer is Option C: One electron is involved in the production of one molecule of CO₂.

According to law of equivalence,

Equivalence of acid = Equivalence of base.

Equivalence of acid = Normality x volume = 0.1 $ \times $ v

As we know base produce OH^$-$ ion, so moles of base is same as moles of OH^$-$ ion = 0.04

Another formula of equivalence = n factor $ \times $ number of moles

$\therefore\,\,\,$ Equivalance of base = n factor of OH^$-$ $ \times $ moles of OH^$-$ = 1 $ \times $ 0.04

As for any ion, the charge of that ion is the n factor of that ion. Here OH^$-$ has 1 negative charge so it's n factor = 1

$\therefore\,\,\,$ 0.1 $ \times $ v = 1 $ \times $ 0.04

$ \Rightarrow $$\,\,\,$ v = 0.4 L

=   0.4 $ \times $ 1000

=    400 ml.

The reducing agent loses electron during redox reaction i.e. oxidises itself.

$\left( 1 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} + 2{H^ + } + 2{e^ - }\,\,\buildrel \, \over \longrightarrow \,\,2{H_2}\mathop O\limits^{ - 2} \,\,({\mathop{\rm Re}\nolimits} d.)$

$\left( 2 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} \,\,\buildrel \, \over \longrightarrow \mathop {{O_2}}\limits^0 + 2{H^ + } + 2{e^ - }\,\,(Ox.)$

$\left( 3 \right)\,\,\,\,\,\,{H_2}\mathop {{O_2}}\limits^{ - 1} + 2{e^ - }\,\,\buildrel \, \over \longrightarrow \,\,2\mathop O\limits^{ - 2} {H^ - }\,\,(Red.)$

$\left( 4 \right)\,\,\,\,\,\,{H_2}O_2^{ - 1} + 2O{H^ - }\,\buildrel \, \over \longrightarrow \mathop {{O_2}}\limits^0 + {H_2}O + 2{e^ - }\,\,\left( {Ox.} \right)$

After balancing the reaction we get

$2MnO_4^- + 5C_2O_4^{2-}$ + 16H⁺ $\to$ 2Mn²⁺ + 10CO₂ + $8{H_2}O$

$\therefore$ $x$ = 2, $y$ = 5 and z = 16

$2H{I^{ - 1}} + {H_2}\mathop {S{O_4}}\limits^{ + 6} \to I_2^0 + \mathop {S{O_2}}\limits^{ + 4} + 2{H_2}O$

in this reaction oxidation number of $S$ is decreasing from $+6$ to $+4$ hence undergoing reduction and for $HI$ oxidation Number of $I$ is increasing from $-1$ to $0$ hence undergoing oxidation therefore ${H_2}S{O_4}$ is acting as oxidising agent.

The oxidation state shows a change only in (d)