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Nernst Equation NEET 2026: Hydrogen Electrode Solution

NEET 2026 Chemistry Redox Reactions and Electrochemistry Nernst Equation

By Founder, JEEnius - IIT Kanpur Alumni · Nov 11, 2026 · 3 min read

Medium 2 min target

Calculate emf of the half cell given below :

Pt (s) | H₂(g, 2 atm) | HCl (aq, 0.02 M)

EH2/H+∘=0 V

(Given: 2.303 RTF=0.059, log 2 = 0.3010)

Show answerAnswer

D) 0.109 V

Explanation

The half-cell reaction is:

H2(g)⟶2H+(aq)+2e−

Using the Nernst equation:

E=E∘−2.303RTnFlog[H+]2PH2

Given:
[H+]=0.02 M (from HCl)
PH2=2 atm
2.303RTF=0.059
n=2

Substituting values:

E=0−0.0592log(0.02)22

E=−0.0295log(0.00042)

E=−0.0295log(2×10−4)

E=−0.0295(log2+log10−4)

E=−0.0295(0.3010−4)

E=−0.0295×(−3.699)

E=0.109 V

Watch the full solution, worked step by step.

What is the answer to the Nernst equation NEET 2026 question?

The supplied official solution gives +0.109 V, option D. In this Nernst equation NEET 2026 calculation, the reaction quotient is below one, so its logarithm is negative. Subtracting that negative term gives the supplied positive answer.

A platinum hydrogen electrode contains hydrogen gas at 2 atm and aqueous HCl at 0.02 M. Calculate its half-cell potential using the supplied standard potential and constants:

E∘=0 V,2.303RTF=0.059,log2=0.3010

The supplied options are:

The question bank labels this medium difficulty, with an expected solving time of 90 seconds. That is a question-bank target, not a measured student average.

How do you write the reaction and build the reaction quotient?

The supplied official solution uses hydrogen ions in the numerator, squared, and hydrogen pressure in the denominator. The square comes from the hydrogen-ion coefficient in the written reaction. The electron count comes from the two electrons, not from the pressure or acid concentration.

The supplied reaction and electron count are:

H2(g)⟶2H+(aq)+2e−

n=2 Under the question’s complete-dissociation treatment of HCl, use:

[H+]=0.02 M

The reaction produces hydrogen ions and consumes hydrogen gas, giving:

Q=[H+]2PH2

Platinum is an inert electrode: it provides the conducting surface but does not appear in the quotient. Electrons are excluded from the quotient as well.

The supplied official working expression is:

E=E∘−0.0592log([H+]2PH2)

This reproduces the supplied reaction-expression pairing, not a universal sign rule for oxidation and reduction potentials. Use the concentration and pressure numerically as supplied; a fuller treatment would use dimensionless activities relative to standard states.

How does the logarithm give positive 0.109 V?

The positive result comes from subtracting a negative logarithm. Square the hydrogen-ion concentration, divide by hydrogen pressure, then evaluate the logarithm. Keep the negative logarithm in brackets during multiplication so that both minus signs remain visible.

Substitute the supplied values:

E=0−0.0592log((0.02)22)

Square before dividing: (0.02)2=0.0004

0.00042=0.0002=2×10−4

Evaluate the coefficient separately, without changing the supplied constant:

0.0592=0.0295
E=−0.0295log(2×10−4)

Expand the base-ten logarithm using the supplied value. The negative exponent contributes a negative term:

log(2×10−4)&=log2+log10−4&=0.3010−4&=−3.6990

Now multiply, retaining both minus signs:

E=−0.0295×(−3.6990)=+0.1091205 V

Round only after multiplication to match the options:

E=+0.109 VOption D

Sign check: the quotient is below one, so its logarithm is negative. In the displayed official expression, subtracting that negative quantity produces a positive potential. This check applies to the expression used here, not to every hydrogen-electrode calculation.

How can reversing the quotient produce option A?

Reciprocating the official quotient while retaining the same prefactor and minus sign produces option A. Start from the quotient already calculated: Q=2×10−4

The inconsistent replacement puts hydrogen pressure in the numerator:

Q′=PH2[H+]2=20.0004=5000

A reciprocal reverses the logarithm’s sign: log(1/Q)=−logQ log5000=+3.6990

Keeping the official expression’s negative prefactor then gives:

Ewrong=−0.0295×3.6990=−0.1091205 V≈−0.109 V

That matches option A. The precise mistake within the supplied calculation is reciprocating the reaction quotient without consistently changing the reaction and potential convention.

A consistently formulated reverse-reaction potential is a separate calculation, not this one-line substitution. Negative hydrogen-electrode potentials are not inherently wrong. Judge the sign from the stated reaction and potential convention, not from whether the answer is positive or negative.

Use this safeguard: write the reaction, identify the electron count, construct the quotient, and only then substitute numbers.

How do you solve the two related Nernst practice questions?

Keep the supplied expression fixed and change only the stated pressure or concentration. These two author-created practice variations from Redox Reactions and Electrochemistry are not additional verified NEET PYQs. Both retain the expression and constants used in the supplied official solution:

E=0−0.0592log([H+]2PH2),2.303RTF=0.059,log2=0.3010

Variation 1: Keep the hydrogen-ion concentration at 0.02 M, but change hydrogen pressure to 1 atm. What is the potential? Calculate the quotient before checking the logarithm below.

Q=(0.02)21=4×10−4
log4=2log2=0.6020
logQ=0.6020−4=−3.3980
E=−0.0295×(−3.3980)=+0.100241 V≈+0.100 V

Variation 2: Keep hydrogen pressure at 2 atm, but change HCl concentration to 0.20 M. Under the same complete-dissociation treatment, what is the potential?

Q=(0.20)22=0.02=2×10−2
logQ=0.3010−2=−1.6990
E=−0.0295×(−1.6990)=+0.0501205 V≈+0.050 V

Compared with the original, increasing hydrogen-ion concentration tenfold makes the quotient one hundred times larger. In the expression used here, that lowers the potential by:

ΔE=−0.0592log100=−0.059 V

Reproduce the original without looking at the working, using its question-bank 90-second target. Then practise mixed questions using Free NEET Chemistry Mock Test: Choose the Right Format.

Frequently asked questions

What is the answer to the Nernst equation NEET 2026 question?

The supplied official solution gives +0.109 V, option D, for hydrogen at 2 atm and HCl at 0.02 M. Its expression gives Q = 0.0002 and log Q = -3.6990, so E = -0.0295 × (-3.6990) = +0.1091205 V. This sign follows the supplied reaction and potential convention, not a universal rule for hydrogen electrodes.

How do I write the reaction quotient for this hydrogen electrode?

For the supplied reaction H₂ → 2H⁺ + 2e⁻, the quotient is Q = [H⁺]²/P(H₂). Using the supplied concentration and pressure numerically gives Q = (0.02)²/2 = 0.0002. Platinum and electrons do not appear in the quotient.

Why is n equal to 2 in this Nernst equation calculation?

The supplied half-reaction H₂ → 2H⁺ + 2e⁻ transfers two electrons, so n = 2. The electron count comes from the balanced reaction, not from the hydrogen pressure or HCl concentration.

Why does reversing the reaction quotient give -0.109 V?

Replacing Q = 0.0002 with its reciprocal, 5000, changes the logarithm from -3.6990 to +3.6990. Retaining the same negative prefactor then gives -0.109 V, option A. Within the supplied calculation, the mistake is reversing the quotient without consistently changing the reaction and potential convention; negative hydrogen-electrode potentials are not inherently wrong.

electrochemistryhydrogen electrodeneet chemistrynernst equationreaction quotient

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