(Equation 1) $F_{2} + 2 e^{-} \to 2 F^{-} (a q)$
$E^{o} = + 2.87 V$
(Equation 2) ${C a}^{2 +} + 2 e^{-} \to C a (s)$
$E^{o} = - 2.76 V$
If we combine the above half reactions in a galvanic cell, which species will be oxidized and which one will reduced ?

F^{-}

will be oxidized and

{C a}^{2 +}

will be reduced

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{C a}^{2 +}

will be oxidized and

F_{2}

will be reduced

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C a (s)

will be oxidized and

F_{2}

will be reduced

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F_{2}

will be reduced and

C a (s)

will be reduced

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Solution

The correct option is C $C a (s)$ will be oxidized and $F_{2}$ will be reduced
As the reduction potential of $F_{2}$ is positive, $F_{2}$ will be reduced and as the reduction potential of $C a^{2 +}$ is negative, $C a (s)$ will be oxidized and $C a^{2 +}$ will be formed after oxidation of $C a (s)$ .

Suggest Corrections

Similar questions

F_{2} + 2 e^{-} \to 2 F^{-} (a q); E = + 2.87 V

C a^{+} + 2 e^{-} \to C a (s); E = - 2.76 V

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

C a^{2 +} + 2 e^{-} \to C a (s); E^{o} = - 2.87 V

M g^{2 +} + 2 e^{-} \to M g (s); E^{o} = - 2.36 V

N i^{2 +} + 2 e^{-} \to N i (s); E^{o} = - 0.25 V

Consider the reaction of water with $F_{2}$ and suggest, in terms of oxidation and reduction, which species are oxidized/reduced.

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