The pressure of H2 required to make the potential of the H2 electrode zero in pure water at 298 K is:

Correct answer: (A) 10-14 atm. Answer: (A) 10-14 atm Solution: For the hydrogen electrode 2H+ + 2e- -> H2(g) with E^° = 0: E = E^° - 0.059/2 PH2[H+]^2 In pure water at 298 K, [H+] = 10-7 M. Setting E = 0 and E^° = 0: 0 = -0.059/2 PH2[H+]^2 PH2[H+]^2 = 1 PH2 = [H+]^2 = (10-7)^2 = 10-14 atm

NEET 2016 Phase 1 · ChemistryNernst equationPrevious Year Question

The pressure of $H X_{2}$ required to make the potential of the $H X_{2}$ electrode zero in pure water at $298 K$ is:

Answer: (A) $1 0^{- 14} atm$ . \textbf{Answer:} (A) $1 0^{- 14} atm$ \textbf{Solution:} For the hydrogen electrode $2 H X^{+} + 2 e X^{-} H X_{2} (g)$ with $E^{\circ} = 0$ : $E = E^{\circ} - \frac{0.059}{2} lo g \frac{P _{H X_{2}}}{[ H X ^{+} ] ^{2}}$ In pure water at $298 K$ , $[H X^{+}] = 1 0^{- 7} M$ .

A. $1 0^{- 14} atm$ ✓
B. $1 0^{- 12} atm$
C. $1 0^{- 10} atm$
D. $1 0^{- 4} atm$

Correct Answer

(A) $1 0^{- 14} atm$

Solution & Explanation

\textbf{Answer:} (A) $1 0^{- 14} atm$ \textbf{Solution:} For the hydrogen electrode $2 H X^{+} + 2 e X^{-} H X_{2} (g)$ with $E^{\circ} = 0$ : $E = E^{\circ} - \frac{0.059}{2} lo g \frac{P _{H X_{2}}}{[ H X ^{+} ] ^{2}}$ In pure water at $298 K$ , $[H X^{+}] = 1 0^{- 7} M$ . Setting $E = 0$ and $E^{\circ} = 0$ : $0 = - \frac{0.059}{2} lo g \frac{P _{H X_{2}}}{[ H X ^{+} ] ^{2}} ⟹ \frac{P _{H X_{2}}}{[ H X ^{+} ] ^{2}} = 1$ $P_{H X_{2}} = [H X^{+}]^{2} = (1 0^{- 7})^{2} = 1 0^{- 14} atm$

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The pressure of HX2​ required to make the potential of the HX2​ electrode zero in pure water at 298 K is:

Solution & Explanation

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The pressure of $H X_{2}$ required to make the potential of the $H X_{2}$ electrode zero in pure water at $298 K$ is: