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Chemistry NEET Questions — Free MCQ Practice

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In which of the following reaction, $ H_2O_2 $ as reducing agent ?

A. $ HNO_2 + H_2O \rightarrow HNO_3 + H_2O $ B. $ 2KMnO_4 + 2H_2O_2 \rightarrow 2MnO_2 + 2KOH + 2H_2O + 3O_2 $ C. $ H_2SO_3 + H_2O_2 \rightarrow H_2SO_4 + H_2O $ D. $ 2Fe^{2+} + H^+ + H_2O_2 \rightarrow 2Fe^{3+} + H^+ + H_2O $

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Explanation

In the given reaction, $2KMnO_4 + 2H_2O_2 ightarrow 2MnO_2 + 2KOH + 2H_2O + 3O_2$, hydrogen peroxide ($H_2O_2$) acts as a reducing agent. This is because it reduces $KMnO_4$ (potassium permanganate) from the +7 oxidation state of manganese in $MnO_4^-$ to the +4 oxidation state in $MnO_2$. Hence, $H_2O_2$ is oxidized to $O_2$ in the process. Therefore, $H_2O_2$ acts as a reducing agent in this reaction.

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Which of the following is a redox reaction ?

A. $ As_2O_3 + 3H_2S \rightarrow As_2S_3 + 3H_2O $ B. $ 2NH_3 + H_2O + CO_2 \rightarrow (NH_4)2CO_3 $ C. $ CaO +3C \rightarrow CO + CaC_2 $ D. $ Be(OH)_2 + 2HCl \rightarrow BeCl_2 + 2H_2O $

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Which of the following is a correct ascending order, when oxidation no. of sulphur of $ H_2SO_3, H_2S_2O_3, H_2S_2O_7, H_2S_2O_8 $ or sulphurus acid, thio sulphuric acid, oleum, dithianic acid is arranged in ascending order ?

A. $ H_2SO_3 \lt H_2S_2O_3 \lt H_2S_2O_7 \lt H_2S_2O_6 $ B. $ H_2SO_3 \lt H_2S_2O_3 \lt H_2S_2O_6 \lt H_2S_2O_7 $ C. $ H_2S_2O_3 \lt H_2SO_3 \lt H_2S_2O_7 \lt H_2S_2O_6 $ D. $ H_2S_2O_3 \lt H_2SO_3 \lt H_2S_2O_6 \lt H_2S_2O_7 $

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Explanation

To determine the oxidation state of sulfur in each compound:

In $H_2SO_3$ (Sulfurous acid), the oxidation state of S is +4.
In $H_2S_2O_3$ (Thiosulfuric acid), it has +2 and +6 oxidation states of sulfur, so the average is +2.
In $H_2S_2O_7$ (Oleum), the oxidation state of S is +6.
In $H_2S_2O_8$ (Peroxydisulfuric acid), the oxidation state of S is +7. Thus, the correct ascending order is $ H_2S_2O_3 ightarrow H_2SO_3 ightarrow H_2S_2O_6 ightarrow H_2S_2O_7 $.

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When 0.25 mole $ I^â€“ oxidised in IO_3 ^ â€“ $ then what coulomb of electric charge relates with the reaction theoretically ?

A. $ 5.79 \times 10 ^ 5 C $ B. $ 4.82500 \times 10 ^5 C $ C. $ 2.895 \times 10 ^ 5 C $ D. $ 1.4475 \times 10 ^ 5 C $

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In the sample of iron oxide, the no. of $ Fe^{2+} $ ion is 90% and no. of is 10% then what is the molecular formula ?

A. $ Fe_{0.905}O $ B. $ Fe_{0.952}O $ C. $ Fe_{0.825}O $ D. $ Fe_{0.857}O $

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The oxidation no. of sulphur in $ Al_2(SO_4)_3 $ is

A. +8 B. +7 C. +5 D. +6

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Explanation

To find the oxidation number of sulfur in $ Al_2(SO_4)_3 $, we need to consider the oxidation states of all the elements in the compound.

Aluminum (Al) has an oxidation state of +3.
Oxygen (O) typically has an oxidation state of -2.

The compound $ Al_2(SO_4)_3 $ can be broken down as follows:

\[ 2 imes (+3) + 3 imes (x + 4 imes (-2)) = 0 \]

Solving for x (the oxidation state of sulfur, S):

\[ 2 imes 3 + 3 imes (x - 8) = 0 \] \[ 6 + 3x - 24 = 0 \] \[ 3x - 18 = 0 \] \[ 3x = 18 \] \[ x = 6 \]

Therefore, the oxidation number of sulfur in $ Al_2(SO_4)_3 $ is +6.

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In a balanced equation, $ aP(S) + bH_2O + CO_2 \rightarrow dH_3PO_4 +CO_2 $ ,oxidation no. of oxygen decreases 30, then c -e= ______

A. 3 B. 1 C. 2 D. 4

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In balanced equation, $ aCu_2S + bNO_3 ^- cH^+ \rightarrow 12Cu^{2+} + eSO_4 ^{2â€“} + fNO + gH_2O $ then what will be the change in oxidation no. and the value of b and g respectively ?

A. 30, 10, 8 B. 121, 20, 16 C. 24, 10, 8 D. 60, 20, 16

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$ Mg |Mg ^ {2+} _{(c_1)} P Ag ^ + _{(c-2)} |Ag $ which of the following Nernst equation is correct for the given electrochemical cell ?

A. $ E ^\circ cell = E cell - { RT \over 2F } .ln { [Mg ^{2+}] \over [Ag^+]^2} $ B. $ E ^\circ cell = E cell - { RT \over F } .ln { [Mg ^{2+}] ^{1/2} \over [Ag^+]} $ C. $ E ^\circ cell = E cell - { RT \over 2F } .ln { [Mg ^{2+}] \over [Ag^+]} $ D. $ E ^\circ cell = E cell - { RT \over 2F } .ln { [Ag ^{+}] ^ 2 \over [Mg^{2+}]} $

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What will be the oxidation potential of $ Pt | H_{2 _{(a)_ (1 bar)}} | H ^+ ( P ^H = 11) half cell 25 ^\circ C $ temperature ?

A. 0.177v B. 0.649v C. 0.282v D. 0.616v

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Explanation

For the half-cell reaction involving hydrogen: \[ Pt | H_2 (1 ext{ bar}) | H^+ (a = 1) \] The oxidation potential can be calculated using the Nernst equation: \[ E = E^ ext{Â°} + rac{0.0591}{n} ext{log} rac{[H^+]}{P_{H_2}} \] Given that $PH = 11$, which means the concentration of $H^+$ is $10^{-11}$. Plugging in the values, we get the oxidation potential as 0.177 V.

Chemistry MCQs for NEET — Practice Questions with Answers

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