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EMF

Electrode pot...

Question

Electrode potentials are:
${Z n}^{0} \to {Z n}^{2 +} + 2 e^{-}; E^{0} = + 0.76 V$
${A u}^{0} \to {A u}^{3 +} + 3 e^{-}; E^{0} = - 1.42 V$
What is the potential of the reaction, if a gold foil is kept in a solution containing zinc ions?

- 1.34

V (no reaction)

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+ 1.34

V (no reaction)

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- 2.18

V (no reaction)

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+ 2.18

V (reaction)

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- 0.66

V (reaction)

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Solution

The correct option is C $- 2.18$ V (no reaction)
$E_{c e l l}^{o} = E_{A u}^{o} - E_{Z n}^{o} = - 1.42 - 0.76 = - 2.18 V$
Since, the cell potential is negative, the reaction is non spontaneous. Hence, no reaction will occur.
$A u^{3 +}$ ions are better oxidizing agent than $Z n^{2 +}$ ions.
Similarly, $Z n$ is better reducing agent than $A u$ .

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Similar questions

Z n \to Z n^{2 +} + 2 e^{-}; E = + 0.76 V

A u \to A u^{3 +} + 3 e^{-}; E = - 1.42 V

Z n^{2 +}

N i C l_{2}

S n B r_{2}

N i^{2 +} (a q) + 2 e^{-} \to N i (s)

S n^{2 +} (a q) + 2 e^{-} \to S n (s)

B r_{2} (a q) + 2 e^{-} \to 2 B r^{-} (a q)

C l_{2} (a q) + 2 e^{-} \to 2 C l^{-} (a q)

{Z n}^{2 +} (a q) + {2 e}^{-} \to Z n (s)

0.76 V

F e^{2 +} + 2 e^{-} \to F e; E^{0} = - 0.440 V

F e^{3 +} + 3 e^{-} \to F e; E^{0} = - 0.036

N O_{3}^{-}

E^{o}

+ 0.96 V

E^{o}

V^{2 +} (aq) + 2 e^{-} \to V E^{0} = - 1.19 V

F e^{3 +} (aq) + 3 e^{-} \to F e E^{0} = - 0.04 V

A u^{3 +} (aq) + 3 e^{-} \to A u E^{0} = + 1.40 V

H g^{2 +} (aq) + 2 e^{-} \to H g E^{0} = + 0.86 V

N O_{3}^{-}

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