On the basis of information available for the reaction- 43 Al - O2-23Al2O3- G - -827 kJ-mol of O2the minimum emf required to carry out an electrolysis of Al2O3 is-Given- 1 F - 96500 C-

The correct option is A 2.14 V Al→ Al^3+ + 3e^ 43 mol Al = 43× 3 mol e^ ≡ 4 mol e^ i.e., n = 4 G = nFE 827× 1000 = 4× ...

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Byju's Answer

Standard XII

Chemistry

EMF

On the basis ...

Question

On the basis of information available for the reaction:
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}; △ G = - 827 k J / m o l$ of $O_{2}$
the minimum emf required to carry out an electrolysis of $A l_{2} O_{3}$ is:
(Given: $1 F = 96500 C)$ .

2.14 V

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4.28 V

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6.42 V

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8.56 V

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Solution

The correct option is A $2.14 V$
$A l \to A l^{3 +} + 3 e^{-}$
$\frac{4}{3} m o l A l = \frac{4}{3} \times 3 m o l e^{-}$
$\equiv 4 m o l e^{-}$
i.e., $n = 4$
$△ G = - n F E$
$- 827 \times 1000 = - 4 \times 96500 \times E$
$E = - 2.14 V$ .

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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

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:

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

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 basic of information availbale from the relation

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 {kJ mol}^{- 1}

O_{2}

, the minimum emf required to carry out electrolysis of

A l_{2} O_{3}

is:

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On the basis of information available for the reaction:43Al+O2→23Al2O3;△G=−827 kJ/mol of O2the minimum emf required to carry out an electrolysis of Al2O3 is:(Given: 1 F=96500 C).

The correct option is A 2.14 VAl→Al3++3e−43mol Al=43×3 mol e−≡4 mol e−i.e., n=4△G=−nFE−827×1000=−4×96500×EE=−2.14 V.

On the basis of information available for the reaction:
$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}; △ G = - 827 k J / m o l$ of $O_{2}$
the minimum emf required to carry out an electrolysis of $A l_{2} O_{3}$ is:
(Given: $1 F = 96500 C)$ .

The correct option is A $2.14 V$
$A l \to A l^{3 +} + 3 e^{-}$
$\frac{4}{3} m o l A l = \frac{4}{3} \times 3 m o l e^{-}$
$\equiv 4 m o l e^{-}$
i.e., $n = 4$
$△ G = - n F E$
$- 827 \times 1000 = - 4 \times 96500 \times E$
$E = - 2.14 V$ .