On the basic of information availbale from the relation 43Al-O2 -23Al2O3- - G -827 kJ mol-1 of O2- the minimum emf required to carry out electrolysis of Al2O3 is-

For 1 mole of O_2, O_2 →23× 3 O^2 so that n=23× 6=4 Hence Δ G= nFE given E=Δ G nF= 827000 4 × 96500= 2.142 V ...

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Byju's Answer

Standard IX

Chemistry

Gibb's Energy and Nernst Equation

On the basic ...

Question

On the basic of information availbale from the relation $\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 {kJ mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out electrolysis of $A l_{2} O_{3}$ is:

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 $2.14 V$
For $1$ mole of $O_{2}$ , $O_{2} \to \frac{2}{3} \times 3 O^{2 -}$
so that $n = \frac{2}{3} \times 6 = 4$
Hence $Δ G = - n F E$ given
$E = \frac{Δ G}{- n F} = \frac{- 827000}{- 4 \times 96500} = 2.142 V$

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Similar questions

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}]

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:

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

Δ G

for the reaction,

\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3},

is -772 kJ mol

^{- 1}

O_{2}

. The minimum E.M.F. in volts required to carry out electrolysis of

A l_{2} O_{3}

is:

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

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On the basic of information availbale from the relation 43Al+O2→23Al2O3, ΔG=−827 kJ mol−1 of O2, the minimum emf required to carry out electrolysis of Al2O3 is:

The correct option is A 2.14 VFor 1 mole of O2, O2→23×3O2− so that n=23×6=4 Hence ΔG=−nFE given E=ΔG−nF=−827000−4×96500=2.142 V

On the basic of information availbale from the relation $\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 {kJ mol}^{- 1}$ of $O_{2}$ , the minimum emf required to carry out electrolysis of $A l_{2} O_{3}$ is:

The correct option is A $2.14 V$
For $1$ mole of $O_{2}$ , $O_{2} \to \frac{2}{3} \times 3 O^{2 -}$
so that $n = \frac{2}{3} \times 6 = 4$
Hence $Δ G = - n F E$ given
$E = \frac{Δ G}{- n F} = \frac{- 827000}{- 4 \times 96500} = 2.142 V$