The maximum amount of electrical energy that can be obtained from one mole of $N_{2} O_{3}$ is :

245.11 k J

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

125.10 k J

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

255.43 k J

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

None of these

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

Open in App

Solution

The correct option is A - $245.11 k J$
Since we have,
$E_{c e l l}^{0} = E_{R}^{0} - E_{L}^{0}$

$E_{c e l l}^{0} = 0.40 - (- 0.87) V = 1.27 V$

$Δ G^{0} = - n F E_{c e l l}^{0}$ = $- 2 \times 96500 \times 1.27 J$

$Δ G^{0} = - 245110 J o r 245.11 k J$

Suggest Corrections

Similar questions

The Edison storage cell is represented as :
Fe(s)/FeO(s)/KOH(aq)/Ni₂O₃ (s)/Ni(s)
The half-cell reactions are:
$N i_{2} O_{3} (s) + H_{2} O (l) + 2 e^{-} \to 2 N i O (s) + 2 O H^{-}; E^{\circ} + 0.40$
$F e O (s) + H_{2} O (l) + 2 e^{-} \to F e (s) + 2 O H^{-}; E^{\circ} = - 0.87$ ,
What is the maximum amount of electrical energy that can be obtained from one mole of Ni₂O₃ ?

Q. The maximum amount of

B a S O_{4}

that can be obtained on mixing 0.5 mole

B a C l_{2}

with 1 mole

H_{2} S O_{4}

is:

Q. The maximum energy of a deuteron coming out a cyclotron is

20 M e V

. The maximum energy of proton that can be obtained from this accelerator is:

Q. Maximum energy can be obtained from oxidation of which respiratory substrate :

Join BYJU'S Learning Program

The maximum amount of electrical energy that can be obtained from one mole of N2O3 is :

The correct option is A -245.11kJSince we have,E0cell=E0R−E0LE0cell=0.40−(−0.87)V=1.27VΔG0=−nFE0cell = −2×96500×1.27JΔG0=−245110Jor245.11kJ

The maximum amount of electrical energy that can be obtained from one mole of $N_{2} O_{3}$ is :

The correct option is A - $245.11 k J$
Since we have,
$E_{c e l l}^{0} = E_{R}^{0} - E_{L}^{0}$

$E_{c e l l}^{0} = 0.40 - (- 0.87) V = 1.27 V$

$Δ G^{0} = - n F E_{c e l l}^{0}$ = $- 2 \times 96500 \times 1.27 J$

$Δ G^{0} = - 245110 J o r 245.11 k J$