For I2+2e−→2I− standard reduction potential=+0.54Volt. For 2Br−→Br2+2e− standard oxidation potential=−1.09Volt. For Fe→Fe2++2e−, standard oxidation potential=+0.44Volt. Which of the following reaction(s) is/are spontaneous:
A
Br2+2I−→2Br−+I2
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B
Fe+Br2→Fe2++2Br−
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C
Fe+I2→Fe2++2I−
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D
I2+2Br−→2I−+Br2
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Solution
The correct options are ABr2+2I−→2Br−+I2 BFe+Br2→Fe2++2Br− CFe+I2→Fe2++2I− (a) Br2+2e−→2Br−EoBr2/Br−=1.09V 2I−→I2+2e−EoI−/I2=−0.54V ∴forBr2+2I−→I2+2Br−Eo=(1.09−0.54)V=0.55V Hence this is a spontaneous reaction.
(b) Fe→Fe2++2e−EoFe/Fe2+=0.44V Br2+2e−→2Br−EoBr2/Br−=1.09V ∴Br2+Fe→Fe2++2Br−Eo=(0.44+1.09)V=1.53V So this is also spontaneous
(c) Fe→Fe2++2e−EoFe/Fe2+=0.44V I2+2e−→2I−EoI2/I−=0.54V ∴I2+Fe→Fe2++2I−Eo=(0.44+0.54)V=0.98V Hence this is also a spontaneous reaction.
(d) I2+2e−→2I−EoI2/I−=0.54V 2Br−→Br2+2e−EoBr−/Br2=−1.09V ∴I2+2Br−→Br2+2I−Eo=(0.54−1.09)V=−0.55V So this reaction is not spontaneous