Given that E H 2 O | H 2| P =0 at 298 K. The pressure of H 2( g ) would be

Question

Given that E H 2 O | H 2| P =0 at 298 K. The pressure of H 2( g ) would be (A) 10-7 bar (B) 10-10 bar (C) 10-12 bar (D) 10-14 bar

Tardigrade Admin · Accepted Answer

The cell reaction is: 2 H ++2 e arrow H 2 E text cell =E text cell °-(0.059/2) log (p H 2/[ H +]2) E cell ° =0, E text cell =0 ∴ p H 2=[ H +]2 In pure water, [ H +]=10-7(M) ∴ p H 2=10-14 bar.

Q. Given that $E_{H_{2} O ∣ H_{2} ∣ P} = 0$ at $298 K$ . The pressure of $H_{2} (g)$ would be

$1 0^{- 7}$ bar

$1 0^{- 10}$ bar

$1 0^{- 12}$ bar

$1 0^{- 14}$ bar

The cell reaction is $: 2 H^{+} + 2 e \to H_{2}$

$E_{cell} = E_{cell}^{\circ} - \frac{0.059}{2} lo g \frac{p _{H_{2}}}{[ H ^{+} ] ^{2}}$

$E_{ce ll}^{\circ} = 0,$

$E_{cell} = 0$

$∴ p_{H_{2}} = [H^{+}]^{2}$

In pure water, $[H^{+}] = 1 0^{- 7} (M)$

$∴ p_{H_{2}} = 1 0^{- 14} ba r .$

Q. Given that EH2​O∣H2​∣P​=0 at 298K. The pressure of H2​(g) would be

10−7 bar

10−10 bar

10−12 bar

10−14 bar