Can someone please explain to me how you figure out how many electrons are transferred in the following reaction.
Fe^2+ + MnO_4 ^- + H^+ ------> Mn^2+ + H_2O + Fe^3+
Fe^2+ + MnO_4 ^- + H^+ ------> Mn^2+ + H_2O + Fe^3+
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But for the manganese in the permanganate ion being reduced to manganese(II), that involves a transfer of 5 electrons. This is why you need five Fe atoms for every MnO4^- ion.
5(Fe2+ --> Fe3+ + 1e-)
8H+ + MnO4^- + 5e- --> Mn2+ + 4H2O
------------------------ -----------------------------
8H+ + 5Fe2+ + MnO4^- -->Mn2+ + 5Fe3+ + 4H2O
Therefore, there are five electrons transferred per mole of reaction.
5(Fe2+ --> Fe3+ + 1e-)
8H+ + MnO4^- + 5e- --> Mn2+ + 4H2O
------------------------ -----------------------------
8H+ + 5Fe2+ + MnO4^- -->Mn2+ + 5Fe3+ + 4H2O
Therefore, there are five electrons transferred per mole of reaction.
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CHANGE IN OXIDATION NUMBER........determines the number of electrons transferred.
Fe^2+--->Fe^3+ + e.........one electron is transferred.
BUT FIRST BALANCE THE EQUATION....as actual number of electrons can only be decided from balance chemical equation.
Fe^2+--->Fe^3+ + e.........one electron is transferred.
BUT FIRST BALANCE THE EQUATION....as actual number of electrons can only be decided from balance chemical equation.