Calculate emf of the half cell given below : Pt (s) | H2 (g, 2 atm) | HCI (aq, 0.02 M) Electrochemistry NEET 2026 PYQ

Calculate emf of the cell given below:
Pt(s)| H2(g, 2 atm)|HCl (aq,0.02 M)
Eo H2/H+= 0V
(Given: 2.303RT/F= 0.059, log2= 0.3010)

For the given reaction: 1/2 H₂ + 1/2 Cl₂  ⟶ HCl + 1 e⁻

 H₂ + Cl₂  ⟶ 2 HCl + 2  e⁻ [ Balanced Equation]

Now the question comes, what will be the value of emf?

To determine the value of emf you just simply have to use the Nernst Equation mentioned in the Chapter Electrochemistry in Class 12 NCERT chemistry book, which is:

Ecell = E⁰cell - 2.303RT/nF Log [product]/[reactant] 

Ecell = - 0.059/2 Log[HCl]²/[H₂] Given, 2.303RT/F = 0.059

Ecell = - 0.059/2  Log[0.02]²/[2] [E⁰cell = 0 V]    [n = No. of electrons] 

Ecell = - 0.059/2  [Log 2 x 10⁻⁴] 

Ecell = - 0.059/2  [Log 2 + Log10⁻⁴]

Ecell = - 0.059/2  [0.3010 - 4]                                                                   

Ecell = - 0.059/2  [-3.7]

Ecell = + 0.109 [option 1 is the correct answer]

Core Concept of this question:  While you are dealing with this kind of  questions from the chapter Electrochemistry especially for NEET you should always look after the half cell reaction given in the question. whether oxidation or reduction is happening and then balance the reaction and you will be good to go.