$Δ G = - 827 k J {m o l}^{- 1}$ of $O_{2}$ the minimum e.m.f required to carry out an electrolysis of ${A l}_{2} O_{3}$ is:
$(F = 96500 C {m o l}^{- 1})$

2.14

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4.28

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6.42

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8.56

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Solution

The correct option is D $2.14$ V
Since ${| W |}_{m i n} = | - n E F | = | - Δ G |$

Or $E = \frac{Δ G}{n F} = (\frac{827 \times 1000}{96500 \times 4}) = 2.14 V$

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Similar questions

The minimum EMF required to carry out the electrolysis of CuO is ( Given ∆G^o for the reaction is - 827 kJ mol^-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}]

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:

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

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:

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ΔG=−827kJmol−1 of O2 the minimum e.m.f required to carry out an electrolysis of Al2O3 is:(F=96500Cmol−1)

The correct option is D 2.14 VSince |W|min=|−nEF|=|−ΔG|Or E=ΔGnF=(827×100096500×4)=2.14V

$Δ G = - 827 k J {m o l}^{- 1}$ of $O_{2}$ the minimum e.m.f required to carry out an electrolysis of ${A l}_{2} O_{3}$ is:
$(F = 96500 C {m o l}^{- 1})$

The correct option is D $2.14$ V
Since ${| W |}_{m i n} = | - n E F | = | - Δ G |$

Or $E = \frac{Δ G}{n F} = (\frac{827 \times 1000}{96500 \times 4}) = 2.14 V$