For a cell reaction-2H2g-O2g- 2H2O l - S0298-0-32 kJK-1- What is the value of - fH0298H20-lGiven - O2g-4H-aq-4e- 2H20l- E0-1-23V

nbsp;2H_2(g)→ 4H^++4e^ ; E=0.0 V O_2(g)+4H^+(aq)+4e^ → 2H_20(l); E=1.23 V 2H_2(g)+O_2(g)→ 2H_2O(l); E^0_cell=1.23V ∴△ G^0_298= nFE^0= 4× 96500× 1.23 = 474. ...

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Byju's Answer

Standard XII

Chemistry

Collision Theory

For a cell re...

Question

For a cell reaction,
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) △ S_{298}^{0} = - 0.32 {kJK}^{- 1}$ . What is the value of $△_{f} H_{298}^{0} (H_{2} 0, l)$
Given : $O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} 0 (l); E^{0} = 1.23 V$

- 285.07 {kJmol}^{- 1}

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- 570.14 {kJmol}^{- 1}

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- 385.07 {kJmol}^{- 1}

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None of these

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Solution

The correct option is D None of these
$2 H_{2} (g) \to 4 H^{+} + 4 e^{-}; E = 0.0 V$
$O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} 0 (l); E = 1.23 V$
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l); E_{c e l l}^{0} = 1.23 V$
$∴△ G_{298}^{0} = - n F E^{0} = - 4 \times 96500 \times 1.23$
$= - 474.78 kJ$
We know that,
$Δ G^{0} = Δ H - T Δ S$
$\Rightarrow - 474.78 = Δ H - 298 \times (- 0.032)$
$\Rightarrow Δ H = - 465.244 kJ/mol$

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Similar questions

Q. For a cell reaction

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l); △_{r} S_{298}^{\circ} = - 0.32 k J / K

. What is the value of

△_{f} H_{298}^{\circ} (H_{2} O, l)

?
Given

O_{2} (g) + 4 H^{+} (a q .) + 4 e^{-} ⟶ 2 H_{2} O (l); E^{\circ} = 1.23 V

Q. The reaction,

2 H_{2} O (l) \to 4 H^{+} (a q .) + O_{2} (g) + 4 e^{-}

is?

Q. During electrolysis,

O_{2} (g)

is evolved at anode in

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) C u (s) ⟶ & C u^{2 +} (a q) + 2 e^{-} E_{C u | C u^{2 +}}^{⊝} = - 0.34 V (w r o n g) 2 H_{2} O (l) ⟶ & O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} E_{H_{2} O | O_{2}}^{⊝} = - 1.23 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

\begin{matrix} I n (d) \end{matrix}

⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ \begin{matrix} (c o r r e c t) N i (s) ⟶ & N i^{2 +} (a q) + 2 e^{-} E_{N i | N i^{2 +}}^{⊝} = - 0.25 V (w r o n g) 4 ⊝ O H (a q) ⟶ & O_{2} (g) + 2 H_{2} O (l) + 4 e^{-} E_{⊝ O H | O_{2}}^{\oplus} = - 0.4 V \end{matrix} ⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

Peroxodisulphate salts (e.g., $N a_{2} S_{2} O_{8}$ ) are strong oxidizing agents used as
bleaching agents for fats, oils, etc.

Given :

$O_{2} (g) + 4 H^{\oplus} (a q) + 4 e^{-} \to 2 H_{2} O (l)$

$E^{⊖} = 1.23 V$

$S_{2} O_{8}^{2 -} (a q) + 2 e^{-} \to S O_{4}^{2 -} (a q)$

$E^{⊖} = 2.01 V$

Which of the following statements is (are) correct ?

Q. The reduction of oxygen gas in

1 M

aqueous acid is represented by this equation (

E^{o} = 1.23 V

)

O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} O (l)

The potential for this reaction at physiological pH, where

[H^{+}] = 1 \times 10^{- 7}

and the partial pressure of

O_{2}

1

atm, will be closest to which value?

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For a cell reaction, 2H2(g)+O2(g)→2H2O(l) △S0298=−0.32 kJK−1. What is the value of △fH0298(H20,l) Given : O2(g)+4H+(aq)+4e−→2H20(l); E0=1.23V

The correct option is D None of these 2H2(g)→4H++4e−; E=0.0 V O2(g)+4H+(aq)+4e−→2H20(l); E=1.23 V 2H2(g)+O2(g)→2H2O(l);E0cell=1.23V ∴△G0298=−nFE0=−4×96500×1.23 =−474.78 kJ We know that, ΔG0=ΔH−TΔS ⇒−474.78=ΔH−298×(−0.032) ⇒ΔH=−465.244 kJ/mol

For a cell reaction,
$2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) △ S_{298}^{0} = - 0.32 {kJK}^{- 1}$ . What is the value of $△_{f} H_{298}^{0} (H_{2} 0, l)$
Given : $O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} 0 (l); E^{0} = 1.23 V$