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Question

# For I2+2e−→2I− standard reduction potential=+0.54 Volt. For 2Br−→Br2+2e− standard oxidation potential=−1.09 Volt. For Fe→Fe2++2e−, standard oxidation potential=+0.44 Volt. Which of the following reaction(s) is/are spontaneous:

A
Br2+2I2Br+I2
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B
Fe+Br2Fe2++2Br
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C
Fe+I2Fe2++2I
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D
I2+2Br2I+Br2
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Solution

## The correct options are A Br2+2I−→2Br−+I2 B Fe+Br2→Fe2++2Br− C Fe+I2→Fe2++2I−(a) Br2+2e−→2Br− EoBr2/Br−=1.09 V 2I−→I2+2e− EoI−/I2=−0.54 V ∴for Br2+2I−→I2+2Br− Eo=(1.09−0.54) V=0.55 V Hence this is a spontaneous reaction. (b) Fe→Fe2++2e− EoFe/Fe2+=0.44 V Br2+2e−→2Br− EoBr2/Br−=1.09 V ∴Br2+Fe→Fe2++2Br− Eo=(0.44+1.09) V=1.53 V So this is also spontaneous (c) Fe→Fe2++2e− EoFe/Fe2+=0.44 V I2+2e−→2I− EoI2/I−=0.54 V ∴I2+Fe→Fe2++2I− Eo=(0.44+0.54) V=0.98 V Hence this is also a spontaneous reaction. (d) I2+2e−→2I− EoI2/I−=0.54 V 2Br−→Br2+2e− EoBr−/Br2=−1.09 V ∴I2+2Br−→Br2+2I− Eo=(0.54−1.09) V=−0.55 V So this reaction is not spontaneous

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