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Electrode Potential

AlCl3 coupled...

Question

$A l C l_{3}$ coupled with normal hydrogen electrode gives an e.m.f. of $1.66 V$ . The standard oxidation electrode potential of aluminium is:

- 1.66 V

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+ 1.66 V

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- 0.83 V

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+ 0.83 V

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Solution

The correct option is A $- 1.66 V$
$E_{c e l l} = E_{R} - E_{L}$
$1.66 = E_{R} - E_{S H E} [∵ E_{S H E} = 0]$
$E_{R} = 1.66$
${R e d}^{n}$ potential i.e. ${A l}^{3 +} + 3 e^{-} \to A l E = 1.66$
$A l \to {A l}^{3 +} + 3 e^{-} E^{0} = - 1.66 Y$

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Similar questions

A l^{3 +} (a q .) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V

C u^{2 +} (a q .) + 2 e^{-} \to C u (s); E^{\circ} = + 0.34 V

E^{θ}

\frac{A l^{+ 3}}{A l}

\frac{T i^{+ 3}}{T i}

M^{3 +}

F e^{3 +} (a q) + e^{-} \to F e^{2 +} (a q); E^{\circ} = + 0.77 V

A l^{3 +} (a q) + 3 e^{-} \to A l (s); E^{\circ} = - 1.66 V

B r_{2} (a q) + 2 e^{-} \to 2 B r^{-}; E^{\circ} = + 1.09 V

E^{0}

Z n / Z n + 2, E^{0} = - 0.76 V

C u^{+ 2} / C u, E^{0} = + 0.34 V

A l / A l^{+ 3}, E^{0} = - 1.66 V

H g^{2 +} / H g, E^{0} = + 0.885 V

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