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Valency and Oxidation State

E∘Fe2+/Fe = -...

Question

$E_{F e^{2 +} / F e}^{\circ} = - 0.44 V, E_{C u^{2 +} / C u}^{\circ} = 0.34 V, E_{A g^{2 +} / g}^{\circ} = 0.80 V$ . Will Fe displace Cu from $C u S O_{4}$ solution? Will Ag displace Cu from $C u S O_{4}$ solution?
1. Yes, Yes
2. No, No
3. Yes, No
4. No, Yes

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Solution

$F e + C u^{2 +} \to F e^{2 +} + C u$
$E^{\circ} = E_{F e / F e^{2 +}}^{\circ} + E_{C u^{2 +} / C u}^{\circ}$
$= 0.44 + 0.34 = 0.78 V$
This reaction (a) is spontaneous
$2 A g + C u^{2 +} \to 2 A g^{+} + C u$
$E^{\circ} = E_{A g / A g^{+}} + E_{C u^{2 +} / C u}$
$= - 0.80 + 0.34$
$= - 0.46 V$
This reaction (b) is not spontaneous.

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Similar questions

F e^{2 +} + 2 e^{-} \to F e

E^{\circ} = - 0.44 V

C u^{2 +} + 2 e^{-} \to C u

E^{\circ} = + 0.34 V

E^{\circ}

F e + C u^{2 +} ⇋ F e^{2} + C u

C u {S O}_{4}

E_{{C u}^{2 +} / C u}^{o} = 0.34 V

E_{{F e / F e}^{2 +}}^{o} = 0.44 V, E_{{A l / A l}^{3 +}}^{o} = 1.66 V

E_{N i / {N i}^{2 +}}^{o} = 0.25 V, E_{{A g}^{+} / A g}^{o} = 0.80 V

Given: $F e_{(s)} \to F e^{2 +} + 2 e^{-}; E^{o} = + 0.44 V$

$P b_{(s)} \to P b^{2 +} + 2 e^{-}; E^{o} = + 0.13 V$

$A g^{+} e^{-} \to A g; E^{o} = + 0.8 V$

$C u^{2 +} 2 e^{-} \to C u; E^{o} = + 0.34 V$

Which of the following metals will oxidise iron?

a) $A g$

b) $C u$

c) $P b$

d) None of these

E_{F e^{2 +} / F e}^{o} = - 0.44, E_{C u^{2 +} / C u}^{o} = 0.34 V, E_{A g + / A g}^{o} = 0.80 V

F e^{2 +} + 2 e^{-} \to F e, E^{\circ} = - 0.44 V

C u^{2 +} + 2 e^{-} \to C u, E^{\circ} = + 0.34 V

A g^{+} + e^{-} \to A g, E^{\circ} = + 0.80 V

F e S O_{4}

C u S O_{4}

C u S O_{4}

A g N O_{3}

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