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Write correctly balanced equation for the following reaction using half reactions:

$bold 1 bold. bold space bold Bi bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold NO subscript bold 3 superscript bold minus bold space bold plus bold space bold H to the power of bold plus bold space bold space bold space bold rightwards arrow bold space bold space bold space bold NO subscript bold 2 bold space bold plus bold space bold Bi to the power of bold 3 bold plus end exponent bold space bold plus bold space bold H subscript bold 2 bold O bold 2 bold. bold space bold I to the power of bold minus bold O subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold O bold space bold space bold space bold rightwards arrow bold space bold space bold space bold I subscript bold 2 bold space bold plus bold space bold OH to the power of bold minus bold 3 bold. bold space bold Cu bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Au to the power of bold plus bold space bold space bold rightwards arrow bold space bold space bold space bold Au bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Cu to the power of bold 2 bold plus end exponent$

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72. Define oxidation number. What are the rules for assigning oxidation number to an element?

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73. Define the term oxidation, reduction, oxidation agent and reducing agent on the basis of oxidation number.

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Short Answer Type

74.
Consider the elements:
Cs, Ne, I and F.

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only positive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither negative nor positive oxidation state.

(i) F (fluorine) exhibits only - ve oxidation state.

(ii) Cs (cesium) exhibits only +ve oxidation state.

(iii) I (iodine) exhibits both +ve and -ve oxidation states.

(iv) (neon) neither exhibits +ve nor - ve oxidation states.

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75. Fluorine reacts with ice and results in the change:
H₂O(s) + F₂(g) → HF(g) + HOF (g)
Justify that this reaction is a redox reaction.

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76. Justify that the reaction

bold space bold space bold 2 bold Na bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold H subscript bold 2 bold left parenthesis bold g bold right parenthesis bold space bold space bold space bold rightwards arrow bold space bold space bold space bold 2 bold NaH bold space bold left parenthesis bold s bold right parenthesis bold space bold is bold space bold a bold space bold redox bold space bold change bold.

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77. Justify that the reaction

bold 2 bold Cu subscript bold 2 bold O bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold Cu subscript bold 2 bold S bold left parenthesis bold s bold right parenthesis bold space bold space bold rightwards arrow bold space bold space bold 6 bold Cu bold left parenthesis bold s bold right parenthesis bold space bold plus bold space bold SO subscript bold 2 bold left parenthesis bold g bold right parenthesis

is a redox reaction. Identify the species oxidised/ reduced, which acts as an oxidant and which acts as a reductant.

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78. Identify the oxidising agent, reducing agent and the substance undergoing oxidation and reduction in the following reactions:

bold space bold space bold left parenthesis bold a bold right parenthesis bold space bold H subscript bold 2 bold SO subscript bold 4 bold space bold plus bold space bold 2 bold HBr bold space bold space bold rightwards arrow bold space bold space bold space bold space bold SO subscript bold 2 bold space bold plus bold space bold Br subscript bold 2 bold space bold plus bold space bold 2 bold H subscript bold 2 bold O bold space bold space bold left parenthesis bold b bold right parenthesis bold space bold 3 bold CuO bold space bold plus bold space bold 2 bold NH subscript bold 3 bold space bold space bold rightwards arrow bold space bold space bold space bold space bold 3 bold Cu bold space bold plus bold space bold N subscript bold 2 bold space bold plus bold space bold 2 bold H subscript bold 2 bold O

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Long Answer Type

79. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)

(b) HCHO(l) + 2[Ag (NH₃)₂]⁺ (aq) + 3OH^-(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O(l)

(c) HCHO (l) + 2 Cu²⁺ (aq) + 5 OH^–(aq) → Cu₂O(s) + HCOO^-(aq) + 3H₂O(l)

(d) N₂H₄(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)

(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

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80. Identify the oxidising agent, reducing agent and the substance undergoing oxidation and reduction in the following reactions:

bold left parenthesis bold a bold right parenthesis bold space bold space bold space bold 2 bold H subscript bold 2 bold S bold space bold plus bold space bold SO subscript bold 2 bold space bold space bold rightwards arrow bold space bold space bold space bold 3 bold S bold plus bold space bold 2 bold H subscript bold 2 bold O bold left parenthesis bold b bold right parenthesis bold space bold 3 bold MnO subscript bold 2 bold space bold plus bold space bold 4 bold Al bold space bold rightwards arrow bold space bold space bold space bold 3 bold Mn bold space bold plus bold space bold 2 bold Al subscript bold 2 bold O subscript bold 3 bold left parenthesis bold c bold right parenthesis bold space bold 2 bold MnO subscript bold 4 superscript bold minus bold space bold plus bold space bold 10 bold Cl to the power of bold minus bold space bold plus bold 16 bold H to the power of bold plus bold space bold space bold rightwards arrow bold space bold space bold space bold space bold 2 bold Mn to the power of bold 2 bold plus end exponent bold space bold plus bold space bold 5 bold Cl subscript bold 2 bold space bold plus bold space bold 8 bold H subscript bold 2 bold O

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