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Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.

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144.

The Mn³⁺ ions is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂ and H⁺ ion. Write a balanced ionic equation for the reaction.

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Short Answer Type

145.

Balance the following redox reactions by ion-electron method:
$MnO subscript 4 superscript minus left parenthesis aq right parenthesis space plus space SO subscript 2 left parenthesis straight g right parenthesis space rightwards arrow space space Mn to the power of 2 plus end exponent left parenthesis aq right parenthesis space plus space HSO subscript 4 superscript minus left parenthesis aq right parenthesis space left parenthesis in space acidic space solution right parenthesis$

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Long Answer Type

146. Balance the following redox reaction by ion-electron method:

stack MnO subscript 4 superscript minus left parenthesis aq right parenthesis space plus space straight I to the power of minus left parenthesis aq with left parenthesis in space basic space medium right parenthesis below right parenthesis space space space rightwards arrow space space space MnO subscript 2 left parenthesis straight s right parenthesis space plus space straight I subscript 2 left parenthesis straight s right parenthesis

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147.
Permanganate (VII) ion, $MnO subscript 4 superscript minus$ in basic solution oxidises iodide ion, I^- to produce molecule iodine (I₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent redox reaction.

The skeleton ionic equation is

1. Write the oxidation and reduction half-reactions by observing the changes in oxidation number and writing these separately.
Oxidation half reaction:
-1 0
I^-(aq)

I₂(s)
Reduction half reaction:
+7 +4

2. Balance the oxidation half reaction
(i) Balance 1 atoms by multiplying I^- by 2
-1 0

(ii) Add 2 electrons towards R.H.S. to balance the charges on iodine atoms

3. Balancing the reduction half reaction
(i) Add 3 electrons towards L.H.S. to balance the charge on Mn

(ii) Balance O atoms by adding two H₂O molecules towards R.H.S.

(iii) Balance H atoms by adding four H⁺ towardS
L.H.S.

Since the reaction takes place in a basic solution, therefore for four H⁺ ions, we add four OH^- ions to both sides of the equation.

[Balanced reduction half reaction]

4. Multiply balanced oxidation half-reaction by 3 and balanced reduction half-reaction by 2 and add both the half reactions.

This is a balanced equation.

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148.

Balance the following equations in basic medium by ion-electron method and oxidation number method:

$straight P subscript 4 left parenthesis straight s right parenthesis space plus space OH to the power of minus left parenthesis aq right parenthesis space rightwards arrow space space PH subscript 3 left parenthesis straight g right parenthesis space plus space HPO subscript 2 superscript minus left parenthesis aq right parenthesis$

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149. Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.

straight N subscript 2 straight H subscript 4 left parenthesis straight l right parenthesis space plus space ClO subscript 3 superscript minus left parenthesis aq right parenthesis space space rightwards arrow space space space NO left parenthesis straight g right parenthesis space plus space Cl to the power of minus left parenthesis straight g right parenthesis

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150. Balance the following equation in basic medium by ion-electron method and oxidation number method and identify the oxidising agent and the reducing agent.