Delta G the reaction- displaystyle frac-4-3- Al - O-2 rightarrow frac-2-3- Al-2O-3- is -772 kJ mol-1- of O-2- The minimum E-M-F- in volts required to carry out electrolysis of Al-2O-3 is-

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Toppr Admin · Accepted Answer

-2-Al-x2192-Al3-3e-x2212-43molofAl-43-xD7-3-xA0-mole-x2212-4-xA0-mol-xA0-e-x2212-n-4Now-x394-G-x2212-nFE-xA0-x2212-772-xD7-1000J-x2212-4-xD7-96500-xD7-E-x2234-E-772-xD7-10004-xD7-96500-2-0V-xA0-

$Δ G$ 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,

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

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,

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