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Question
Can Fe3+ oxidises bromide to bromine under standard conditions?
Given: \[\ce{E^0_{{Fe^{3+}|Fe^{2+}}}}\] = 0.771 V
\[\ce{E^0_{{Br_{2}|Br^-}}}\] = −1.09 V
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Solution
Required half cell reaction
\[\ce{2Br^- -> Br2 + 2e^-}\] `("E"_"ox"^0)` = −1.09 V
\[\ce{2Fe^{3+} + 2e^- -> 2Fe^{2+}}\] `("E"_"red"^0)` = +0.771 V
\[\ce{2Fe^{3+} + 2Br^- -> 2Fe^{2+} + Br2}\] `("E"_"cell"^0)` = ?
`"E"_("cell")^0 = ("E"_("ox")^0) + ("E"_("red")^0)`
= −1.09 V + 0.771
= −0.319 V
\[\ce{E^0_{cell}}\] is – ve; ΔG is +ve and the cell reaction is non spontaneous. Hence Fe3+ cannot oxidises Br− to Br2.
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Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half-reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half-cells. The reading of the voltmeter gives the cell voltage or cell potential or electromotive force. If \[\ce{E^0_{cell}}\] is positive the reaction is spontaneous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu2+. This was first formulated by Faraday in the form of laws of electrolysis.
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