$Δ G$ for the reaction, $\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3},$ is -772 kJ mol $^{- 1}$ of $O_{2}$ . The minimum E.M.F. in volts required to carry out electrolysis of $A l_{2} O_{3}$ is:

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Solution

(2)
$A l \to A l^{3 +} + 3 e^{-}$
$\frac{4}{3} m o l o f A l = \frac{4}{3} \times 3 m o l e^{-} = 4 m o l e^{-}$
$n = 4$

Now,

$Δ G = - n F E$
$- 772 \times 1000 J = - 4 \times 96500 \times E$
$∴ E = \frac{772 \times 1000}{4 \times 96500} = 2.0 V$

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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:

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

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 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:

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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ΔG for the reaction, 43Al+O2→23Al2O3, is -772 kJ mol−1 of O2. The minimum E.M.F. in volts required to carry out electrolysis of Al2O3 is:

(2)Al→Al3++3e−43molofAl=43×3 mole−=4 mol e−n=4Now,ΔG=−nFE −772×1000J=−4×96500×E∴E=772×10004×96500=2.0V

$Δ G$ for the reaction, $\frac{4}{3} A l + O_{2} \to \frac{2}{3} A l_{2} O_{3},$ is -772 kJ mol $^{- 1}$ of $O_{2}$ . The minimum E.M.F. in volts required to carry out electrolysis of $A l_{2} O_{3}$ is:

(2)
$A l \to A l^{3 +} + 3 e^{-}$
$\frac{4}{3} m o l o f A l = \frac{4}{3} \times 3 m o l e^{-} = 4 m o l e^{-}$
$n = 4$

Now,

$Δ G = - n F E$
$- 772 \times 1000 J = - 4 \times 96500 \times E$
$∴ E = \frac{772 \times 1000}{4 \times 96500} = 2.0 V$