On the basis of information available from the reaction-4-3Al-O2 -2-3Al2O3 - G -827 KJ-mol of O2- The minimum e-m-f- required to carry outelectrolysis of Al2O3 is

For 1 mole of O_2 O_2 →2/3× 3O^2 n=4 G^∘= n FE^∘ cell E^∘= Δ G^∘/nf= 827000/4 × 96500=2.14V ...

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Byju's Answer

Standard XI

Chemistry

Gibb's Energy and Nernst Equation

On the basis ...

Question

On the basis of information available from the reaction:
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3} Δ G = - 827 K J / m o l$ of $O_{2}$ . The minimum e.m.f. required to carry out
electrolysis of $A l_{2} O_{3}$ is

2.14V

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4.28V

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6.42V

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8.56V

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Solution

The correct option is A
2.14V

For 1 mole of $O_{2}$
$O_{2} \to \frac{2}{3} \times 3 O^{2 -} n = 4 G^{\circ} = - n F E^{\circ} c e l l E^{\circ} = \frac{- Δ G^{\circ}}{n f} = \frac{- 827000}{4 \times 96500} = 2.14 V$

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Similar questions

On the basis of information available from the reaction:
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3} Δ G = - 827 K J / m o l$ of $O_{2}$ . The minimum e.m.f. required to carry out
electrolysis of $A l_{2} O_{3}$ is

Q. On the basis of information available from the reaction:

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3} Δ G = - 827 K J / m o l

O_{2}

. The minimum e.m.f. required to carry out
electrolysis of

A l_{2} O_{3}

Q. On the basis of the information available from the reaction:

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 k J m o l^{- 1}

O_{2}

, the minimum emf required to carry out the electrolysis of

A l_{2} O_{3}

[F = 96500 C m o l^{- 1}]

Q. On the basis of information available for the reaction:

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}; △ G = - 827 k J / m o l

O_{2}

the minimum emf required to carry out an electrolysis of

A l_{2} O_{3}

is:
(Given:

1 F = 96500 C)

On the basis of information available from the reaction

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 k J . m o l^{- 1}

The minimum emf required to carry out the electrolysis of $A l_{2} O_{3}$ is:

$[F = 96500 C m o l^{- 1}]$

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On the basis of information available from the reaction: 43Al+O2→23Al2O3ΔG=−827 KJ/mol of O2. The minimum e.m.f. required to carry out electrolysis of Al2O3 is

The correct option is A 2.14V For 1 mole of O2 O2→23×3O2− n=4G∘=−n FE∘ cellE∘=−ΔG∘nf=−8270004×96500=2.14V

On the basis of information available from the reaction:
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3} Δ G = - 827 K J / m o l$ of $O_{2}$ . The minimum e.m.f. required to carry out
electrolysis of $A l_{2} O_{3}$ is

The correct option is A
2.14V

For 1 mole of $O_{2}$
$O_{2} \to \frac{2}{3} \times 3 O^{2 -} n = 4 G^{\circ} = - n F E^{\circ} c e l l E^{\circ} = \frac{- Δ G^{\circ}}{n f} = \frac{- 827000}{4 \times 96500} = 2.14 V$