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Electrochemistry question

2010 · Shift 0 · Q20

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Electrochemistry
2010 · Shift 0 · Q20

Electrochemistry question

2010 · Shift 0 · Q20

JEE MainChemistryElectrochemistryMCQ+4 / −1

The Gibbs energy for the decomposition of

Al_2O_3

at 500oC is as follows :

{2 \over 3}A{l_2}{O_3}\to{4 \over 3}Al + {O_2}

,

{\Delta _r}G

= + 966 kJ mol–1 The potential difference needed for electrolytic reduction of

Al_2O_3

at 500oC is at least :

A
4.5 V
B
3.0 V
C
2.5 V
D
5.0 V

View written solutionFree

Correct answer: C

Use the relation between Gibbs energy and cell potential

For an electrochemical process,

$\Delta G = -n F E$

For an electrolytic decomposition, the minimum external potential required is given by the magnitude:

$E_{\min} = \frac{\Delta G}{nF}$

where:

$\Delta G = 966\ \text{kJ mol}^{-1} = 966000\ \text{J mol}^{-1}$
$F = 96500\ \text{C mol}^{-1}$
$n$ = number of moles of electrons involved in the reaction.

Find $n$ for the given decomposition reaction

Given:

$\frac{2}{3}Al_2O_3 \rightarrow \frac{4}{3}Al + O_2$

In $Al_2O_3$ , aluminium is in oxidation state $+3$ and gets reduced to $Al(0)$ .

Each $Al^{3+}$ requires $3$ electrons.

Number of aluminium atoms produced = $\frac{4}{3}$ mol.

So total electrons required:

$n = \frac{4}{3} \times 3 = 4$

Calculate the minimum potential

$E_{\min} = \frac{966000}{4 \times 96500}$

$E_{\min} \approx \frac{966000}{386000} \approx 2.50\ \text{V}$

Match with options

The required potential difference is at least:

$\boxed{2.5\ \text{V}}$

So the correct option is C.

Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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