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

Correct answer: (D) 0.109 V. Answer: (D) 0.109 V Solution: For the hydrogen electrode the Nernst equation (NEET official key convention) is written with reaction quotient Q = [H+]^2PH2 and n = 2: E = E^° - 0.0592 [H+]^2PH2 With [H+] = 0.02 M (from HCl) and PH2 = 2 atm: E = 0 - 0.0295 (0.02)^22 = -0.0295 4 10-42 = -0.0295 (2 10-4) E = -0.0295,( 2 - 4) = -0.0295,(0.301 - 4) = -0.0295×(-3.699) E = +0.109 V, which is option (D).

NEET 2026 (Phase 1) · ChemistryNernst equationPrevious Year Question

Calculate the emf of the half cell given below: $Pt (s) ∣ H X_{2} (g, 2 atm) ∣ HCl (aq, 0.02 M)$ $E_{H X^{+} / H X_{2}}^{\circ} = 0 V$ (Given: $\frac{2.303 R T}{F} = 0.059$ , $lo g 2 = 0.3010$ )

Answer: (D) $0.109 V$ . \textbf{Answer:} (D) $0.109 V$ \textbf{Solution:} For the hydrogen electrode the Nernst equation (NEET official key convention) is written with reaction quotient $Q = \frac{[ H X ^{+} ] ^{2}}{P _{H X_{2}}}$ and $n = 2$ : $E = E^{\circ} - \frac{0.059}{2} lo g \frac{[ H X ^{+} ] ^{2}}{P _{H X_{2}}}$ With $[H X^{+}] = 0.02 M$ (from $HCl$ ) and $P_{H X_{2}} = 2 atm$ : $E = 0 - 0.0295 lo g \frac{( 0.02 ) ^{2}}{2} = - 0.0295 lo g \frac{4 \times 1 0 ^{- 4}}{2} = - 0.0295 lo g (2 \times 1 0^{- 4})$ $E = - 0.0295 (lo g 2 - 4) = - 0.0295 (0.301 - 4) = - 0.0295 \times (- 3.699)$ $E = + 0.109 V$ , which is option (D).

A. $- 0.109 V$
B. $0.035 V$
C. $- 0.035 V$
D. $0.109 V$ ✓

Correct Answer

(D) $0.109 V$

Solution & Explanation

\textbf{Answer:} (D) $0.109 V$ \textbf{Solution:} For the hydrogen electrode the Nernst equation (NEET official key convention) is written with reaction quotient $Q = \frac{[ H X ^{+} ] ^{2}}{P _{H X_{2}}}$ and $n = 2$ : $E = E^{\circ} - \frac{0.059}{2} lo g \frac{[ H X ^{+} ] ^{2}}{P _{H X_{2}}}$ With $[H X^{+}] = 0.02 M$ (from $HCl$ ) and $P_{H X_{2}} = 2 atm$ : $E = 0 - 0.0295 lo g \frac{( 0.02 ) ^{2}}{2} = - 0.0295 lo g \frac{4 \times 1 0 ^{- 4}}{2} = - 0.0295 lo g (2 \times 1 0^{- 4})$ $E = - 0.0295 (lo g 2 - 4) = - 0.0295 (0.301 - 4) = - 0.0295 \times (- 3.699)$ $E = + 0.109 V$ , which is option (D).

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Calculate the emf of the half cell given below: Pt(s) ∣ HX2​(g,2atm) ∣ HCl(aq,0.02M) EHX+/HX2​∘​=0 V (Given: F2.303RT​=0.059, log2=0.3010)

Solution & Explanation

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Calculate the emf of the half cell given below: $Pt (s) ∣ H X_{2} (g, 2 atm) ∣ HCl (aq, 0.02 M)$ $E_{H X^{+} / H X_{2}}^{\circ} = 0 V$ (Given: $\frac{2.303 R T}{F} = 0.059$ , $lo g 2 = 0.3010$ )