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

The correct option is A 2.14 V 4/3 Al +O_2 → 2/3Al_2O_3, Δ G = 827KJ mol^ 1 Δ G = nFE^0 #160; #160; #160; 827 × 10^3J = 2 × E^0 ...

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Byju's Answer

Standard XII

Chemistry

Equilibrium Constant from Nernst Equation

On the basis ...

Question

On the basis of the information available from the reaction:

$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 k J m o l^{- 1}$ of $O_{2}$ , the minimum emf required to carry out the electrolysis of $A l_{2} O_{3}$ is $[F = 96500 C m o l^{- 1}]$ :

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 $2.14 V$
$4 / 3 A l + O_{2} \to 2 / 3 A l_{2} O_{3}, Δ G = - 827 K J m o l^{- 1}$

$Δ G = - n F E^{0}$

$- 827 \times 10^{3} J = - 2 \times E^{0} \times 96500$

$E^{o} = \frac{827 \times 10^{3}}{2 \times 96500}$

$E^{o} = 4.28 V$

The minimum emf required to carry out the electrolysis of $A l_{2} O_{3}$ = $\frac{4.28}{2}$ = $2.14 V$

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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 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)

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}

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On the basis of the information available from the reaction:43Al+O2→23Al2O3, ΔG=−827 kJ mol−1 of O2, the minimum emf required to carry out the electrolysis of Al2O3 is [F=96500 C mol−1]:

The correct option is A 2.14 V4/3Al+O2→2/3Al2O3,ΔG=−827KJmol−1ΔG=−nFE0 −827×103J=−2×E0×96500Eo=827×1032×96500Eo=4.28V The minimum emf required to carry out the electrolysis of Al2O3 = 4.282=2.14V

On the basis of the information available from the reaction:

$\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3}, Δ G = - 827 k J m o l^{- 1}$ of $O_{2}$ , the minimum emf required to carry out the electrolysis of $A l_{2} O_{3}$ is $[F = 96500 C m o l^{- 1}]$ :