Class 12 Chemistry Previous Year Question Paper 2026 (Set – 1 & 2) with Solutions

CBSE Students who are preparing for upcoming 12th chemistry exam must practice this class 12 chemistry solved PYQ 2026. This will be very helpful for students in their exam preparation. Students can understand the exam pattern, question difficulty and also can do self analysis with the help of this PYQ.

We have also given solutions of each question so that student can cross check their answer. Practising Class 12 chemistry Previous Year Question papers 2026 before exam can boost confidence in students and students can score better marks in exam. Different set of PYQ has given in this post and CBSE experts have prepared solutions of this PYQ.

Class 12 chemistry PYQ 2026 SET 1 Question
Class 12 chemistry PYQ 2026 SET 1 Solution
Class 12 chemistry PYQ 2026 SET 2 Question
Class 12 chemistry PYQ 2026 SET 2 Solution

Class 12 chemistry PYQ 2026 SET 1 Question

Class 12 Chemistry PYQ 2026 SET 1 Solution

Class 12 chemistry PYQ 2026 SET 2 Question

Class 12 Chemistry PYQ 2026 SET 2 Solution

Class 12 Chemistry PYQ 2025
Class 12 Chemistry PYQ 2019

CBSE exam is usually held between feb and April every year. And for many students, this exam decides their career specially who want to pursue a career in Chemistry. Practising only chapter is not enough for scoring good marks in exam. Solving Class 12 Chemistry PYQs is also important.

We hope students found this class 12 chemistry solved PYQ 2026 helpful for them. Keep learning if you have any questions you can ask in comment section. Also you can also help your friend by sharing this post to them

Important Question of Class 12 Chemisty PYQ 2026

Question :

Differentiate between the following :

(i) Fibrous protein and Globular protein

(ii) Peptide linkage and Phosphodiester linkage

Soution :

(i) Fibrous protein and Globular protein

Fibrous protein:

  • When polypeptide chains run parallel and are held together by hydrogen and disulphide bonds, then fibre-like structure is formed.
  • It is insoluble in water.

Globular protein:

  • When polypeptides coil around to give a spherical shape.
  • It is soluble in water.

(ii) Peptide linkage and Phosphodiester linkage

Peptide linkage:

  • It is an amide linkage formed between two amino acids through –CONH–.
  • It is formed between –COOH and –NH₂ groups of two amino acids.

Phosphodiester linkage:

  • In phosphodiester linkage, two nucleotides join together.

Question :

Calculate emf and ΔG for the following cell at 298 K

Mg(s)/Mg2+(0.01 M)//Ag+(0.001 M)/Ag(s)\text{Mg(s)} / \text{Mg}^{2+}(0.01\text{ M}) // \text{Ag}^+(0.001\text{ M}) / \text{Ag(s)}
Given : EMg2+/Mg=2.37 VEAg+/Ag=+0.80 V\text{Given : } \text{E}^\circ_{\text{Mg}^{2+}/\text{Mg}} = -2.37\text{ V} \quad \text{E}^\circ_{\text{Ag}^+/\text{Ag}} = +0.80\text{ V}
[1 F=96500 C mol1,log10=1][1\text{ F} = 96500\text{ C mol}^{-1}, \log 10 = 1]

Soution :

Ecell=Ecell0.0592log[Mg2+][Ag+]2\text{E}_{\text{cell}} = \text{E}^\circ_{\text{cell}} – \frac{0.059}{2} \log \frac{[\text{Mg}^{2+}]}{[\text{Ag}^+]^2}
=[0.80+2.37]0.0592log[0.01][0.001]2= [0.80 + 2.37] – \frac{0.059}{2} \log \frac{[0.01]}{[0.001]^2}
=3.170.0592log104= 3.17 – \frac{0.059}{2} \log 10^4
=3.170.118=3.052VΔG=nFEcell= 3.17 – 0.118 = 3.052\text{V} \Delta\text{G} = -n\text{F}\text{E}_{\text{cell}}
=2×96500×3.052= -2 \times 96500 \times 3.052
=589036 Jmol1= -589036\text{ Jmol}^{-1}
=Or 589.036 kJ mol1= \text{Or } -589.036\text{ kJ mol}^{-1}

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