At 25-C- - HfH2O- I-56700 J-mol and energy of ionization of H2OI-19050 J-molIf the reversible EMF at 25-C of the cell is 414-10-x- then the value of x is - Pt - H2g1 atm - H- - -OH- - O2g1 atm -Pt- if at 26-C the emf increase by 0-001158 V-Ans in V

H_2+1/2O_2→ H_2O #160; Δ H_1= 56700 J/mole ...(i) 2H_2O→ 2H^++2OH^ Δ H_2=2× 19050 ....(ii)call reaction :at anode : H_2→ 2H^++2e^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Electron Gain Enthalpy

At 25∘C, Δ ...

Question

At $25^{\circ} C, Δ H_{f} (H_{2} O, I) = - 56700 J / m o l$ and energy of ionization of $H_{2} O (I) = 19050 J / m o l$ If the reversible EMF at $25^{\circ} C$ of the cell is $414 \times 10^{- x}$ , then the value of x is ?
$P t | H_{2} (g) (1 a t m) | H^{+} | | O H^{-} | O_{2} (g) (1 a t m) | P t$ , if at $26^{\circ} C$ the emf increase by 0.001158 V.
Ans in V

Open in App

Solution

Suggest Corrections

Similar questions

25^{o} C

H_{2} O (l)

- 61076 c a l / m o l

H^{+}

O H^{-}

19050 c a l / m o l

25^{o} C

H_{2} (g) (1 a t m) | H^{+} | | O_{2} (g) (1 a t m) | O H^{-}

M | M^{n +} (0.2 M) | | H^{+} (1 M) | H_{2} (g) (1 a t m)

25^{\circ} C

P t | H_{2} (1 a t m) | H C l (0.01 M) | A g C l (s) | A g (s) a t 298 K

Δ G_{t}^{o}

^{o}

- 109.56 \frac{k J}{m o l}

- 130.79 \frac{k J}{m o l}

^{+}

^{-}

P t, H_{2 (g)} (1 atm) / H^{+}, p H = x | |

0.67 V

25^{\circ} C

- 0.28 V

p H

\frac{2.303 R T}{F} = 0.0591 V

M | M^{n +} (0.02 M) | | H^{+} (1 M) | H_{2} (g) (1 atm), P t

25^{\circ} C

0.81 V

0.76 V

Join BYJU'S Learning Program