Hydrazine can be used in fuel cell N2H4aq - O2g- N2g - 2H2OlIf - G0 for this reaction is -600 kJ- what will be the E0 for the cell-

The correct option is C 1.55 V 2N_2H_4_(aq)+O_2_(g)⟶0N_2_(g)+2H_2O_(l) Electrons transferred= 4/N_2H_4 ⟶ N_2+4e^ Δ G^o= nFE^o ⇒ 600 × 10 ...

166

You visited us 166 times! Enjoying our articles? Unlock Full Access!

Work Energy Theorem

Hydrazine can...

Question

Hydrazine can be used in fuel cell
$N_{2} H_{4 (a q)} + O_{2 (g)} \to N_{2 (g)} + 2 H_{2} O_{(l)}$
If $Δ G^{0}$ for this reaction is -600 kJ, what will be the $E^{0}$ for the cell?

1.25 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.85 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1.55 V

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.75 V

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

2 H_{2} (g) + O_{2} (g) ⟶ 2 H_{2} O (l)

Δ G_{f}^{o} (H_{2} O) = - 237.2 k J {m o l}^{- 1}

Δ H_{f}^{0}

H_{2} O_{2}

H_{2} O

2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)

2 H_{2} (g) + O_{2} (g) \to 2 H_{2} O (l) △ S_{298}^{0} = - 0.32 {kJK}^{- 1}

△_{f} H_{298}^{0} (H_{2} 0, l)

O_{2} (g) + 4 H^{+} (a q) + 4 e^{-} \to 2 H_{2} 0 (l); E^{0} = 1.23 V

H_{2} O_{2} (l),

N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) \to N_{2} (g) + 4 H_{2} O (l);

△_{r} H_{1}^{o} = - 818 k J / m o l

N_{2} H_{4} (l) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (l);

△_{r} H_{2}^{o} = - 622 k J / m o l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △_{f} H_{3}^{o} = - 285 k J / m o l

Δ H_{f}^{o}

H_{2} O_{2}

H_{2} O

2 H_{2} O_{2} (l) \to 2 H_{2} O (l) + O_{2} (g)

Join BYJU'S Learning Program