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Electrochemical Series

Which substan...

Question

Which substance is most easily reduced?

I_{2}

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C l_{2}

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F_{2}

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B r_{2}

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A t_{2}

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Solution

The correct option is B $F_{2}$
Fluorine $F_{2}$ is most easily reduced.
Fluorine is most electronegative element and hence, easily accepts an electron to form fluoride ion. This makes fluorine the strongest oxidising agent. It is due to smaller size, low bond enthalpy, high electron affinity and high exothermic hydration energy of the small fluoride ion.

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Similar questions

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

F_{2} > C l_{2} > B r_{2} > I_{2} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

H O C l > H C l O_{2} > H C l O_{3} > H C l O_{4} :

(x^{2} + y^{2})

x

y

I_{2}, {B r}_{2}, {C l}_{2}, F_{2}

I_{2} < {B r}_{2} < {C l}_{2} < F_{2}

F_{2}, C l_{2}, B r_{2}, I_{2}

B r_{2}

N a F, N a C l & N a I

F_{2} > C l_{2} > B r_{2} > I_{2}

F_{2} > C l_{2} > B r_{2} > I_{2}

C l > F > B r > I

F > C l > B r > I

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