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Electrometallurgy

Eo values of ...

Question

$E^{o}$ values of three metals are listed given.
$Z n_{(a q)}^{2 +} + 2 e^{-} \to Z n_{(s)};$ $E^{o} = - 0.76$ V
$F e_{(a q)}^{2 +} + 2 e^{-} \to 2 F e_{(s)};$ $E^{o} = - 0.44$ V
$S n_{(a q)}^{2 +} + 2 e^{-} \to S n_{(S)};$ $E^{o} = - 0.14$ V
Which of the following statements are correct on the basis of the given information?
(i) Zinc will be corroded in preference to iron if zinc coating is broken on the surface.
(ii) If iron is coated with tin and the coating is broken on the surface then iron will be corroded.
(iii) Zinc is more reactive than iron but tin is less reactive than iron.

(i) and (ii) only

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(ii) and (iii) only

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(i), (ii) and (iii)

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(i) and (iii) only

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Solution

The correct option is A (i), (ii) and (iii)
The given are the reduction potentials of the elements. We can notice that,
Iron coated with zinc does not get rusted even if cracks appear on the surface because $Z n$ will take part in redox reaction not $F e$ as $Z n$ is more reactive than $F e$ . It is called Sacrificial method of prevention of rusting.
If iron is coated with tin and cracks appear on the surface. $F e$ will take part in redox reaction because $S n$ is less reactive than $F e$ .

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Similar questions

F e_{(a q)}^{3 +} + e^{-} \to F e_{(a q)}^{2 +}

E^{o} = + 0.77

A l_{(a q)}^{3 +} + 3 e^{-} \to A l_{s}; E^{o} = - 1.66

B r_{2 (a q)} + 2 e^{-} \to 2 B r_{(a q)}^{-}; E^{o} = + 1.66

Z n^{2 +} (a q) + 2 e^{-} ⇌ Z n (s); E^{o} = - 0.762 V

C r^{3 +} (a q) + 3 e^{-} ⇌ C r (s); E^{o} = + 0.762 V

2 H^{+} (a q) + 2 e^{-} ⇌ H_{2} (g); E^{o} = + 0.00 V

F e^{3 +} (a q) + e^{-} ⇌ F e^{2 +} (a q); E^{o} = + 0.770 V

N O_{3}^{-}

E^{o}

0.96 V

E^{o}

V^{2 +} (a q) + 2 e^{-} \to V; E^{o} = - 1.19 V F e^{3 +} (a q) + 3 e^{-} \to F e; E^{o} = - 0.04 V A u^{3 +} (a q) + 3 e^{-} \to A u; E^{o} = 1.40 V H g^{2 +} (a q) + 2 e^{-} \to H g; E^{o} = + 0.86 V

N O_{3}^{-}

(E^{o})

Z n = Z n^{2 +} + 2 e^{-}

E^{o} = + 0.76

F e = F e^{2 +} + 2 e^{-}

E^{o} = + 0.41

F e^{2 +} + Z n \to Z n^{2 +} + F e

Food cans are coated with tin and not with zinc because

(a) zinc is costlier than tin.

(b) zinc has a higher melting point than tin.

(d) zinc is less reactive than tin.

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