Previous Year Papers

Download Solved Question Papers Free for Offline Practice and view Solutions Online.

Test Series

Take Zigya Full and Sectional Test Series. Time it out for real assessment and get your results instantly.

Test Yourself

Practice and master your preparation for a specific topic or chapter. Check you scores at the end of the test.

Balance the following equation by oxidation number method:
$Fe to the power of 2 plus end exponent space plus space Cr subscript 2 straight O subscript 7 superscript 2 minus end superscript space plus space straight H to the power of plus space space rightwards arrow space space Fe to the power of 3 plus end exponent space plus space 2 Cr to the power of 3 plus end exponent space plus space 3 straight H subscript 2 straight O$

213 Views

132.

Balance the following equation by oxidation number method:
$left enclose CrO subscript 4 right parenthesis to the power of 2 minus end exponent space plus space space left parenthesis SO subscript 3 right parenthesis to the power of 2 minus end exponent space space space rightwards arrow space space end enclose open vertical bar Cr left parenthesis OH right parenthesis subscript 4 close vertical bar to the power of 1 minus end exponent space plus space left parenthesis SO subscript 4 right parenthesis to the power of 2 minus end exponent$
(in alkaline medium)

125 Views

133.

Balance the following redox reaction by oxidation number method:

$left parenthesis Cr left parenthesis OH right parenthesis subscript 4 superscript minus space plus space straight H subscript 2 straight O subscript 2 space space rightwards arrow space space space space CrO subscript 4 superscript 2 minus end superscript space plus space straight H subscript 2 straight O space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis Basic space medium right parenthesis$

112 Views

134.

Balance the following equation:
$Cr left parenthesis OH right parenthesis subscript 3 space plus space IO subscript 3 superscript minus space space space rightwards arrow space space space CrO subscript 4 superscript 2 minus end superscript space plus space straight I to the power of minus space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis basic space medium right parenthesis space space space space space space space space$

108 Views

135. Write the net ionic equation for the reaction of potassium dichromate (VI), K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium (III) ion and the sulphate ion.

1750 Views

136.

With the net ionic equation for the reaction of potassium dichromate (VI), K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium (III) ion and the sulphate ion.

148 Views

137. Write the various steps for balancing redox equation by ion-electron method (Half reaction method).

194 Views

138.
Balance the equation by half reaction method:

$straight I subscript 2 space plus space OH to the power of minus space space space space space rightwards arrow space space space space IO subscript 3 superscript minus space plus space straight I to the power of minus space plus space straight H subscript 2 straight O$

1. Write the oxidation and reduction half-reactions by observing the changes in oxidation numbers and write these separately.
0 +5
Oxidation half-reaction:

0
Reduction half-reaction:

2. Balancing the oxidation half reaction.
(i) Balance 1 atoms by multiplying

by 2.
0 +5

(ii) Add 10 electrons towards R.H.S. to balance the charges on iodine atoms,
0 +5

(iii) Since the medium is alkaline, 6H₂O molecules are added towards R.H.S. and 12 OH^-ions towards L.H.S. in order to balance oxygen and hydrogen atoms.

3. Balancing the reduction half reaction.
(i) Balancing 1 atoms by multiplying I^-by2. 0 +5

(ii) Add two electrons towards L.H.S. in order to balance the charges on iodine atoms

4. Multiplying reduction half-reaction by 5 to equate the electrons and add both the half reactions.

This is a balanced redox reaction.

107 Views

139.

Balance the following equation by half reaction method in acidic medium:
$MnO subscript 4 superscript minus space plus space Br to the power of minus space space space rightwards arrow space space space Mn to the power of 2 plus end exponent space plus space Br squared$