Answer to Question #159372 in General Chemistry for Andrea Nicole Billardo

Question

Balance the reaction in acid and base medium using half reaction method

Cr2O7-2 + Fe+2 - Cr+3 + Fe+3

Answer 1

Solution:

$Cr^{6+}+3e^-\rightarrow Cr^{3+}\\Fe^{2+}\rightarrow e^- +Fe^{3+}\\Cr_2O_7^{-2}+6e^-\rightarrow 2Cr^{3+}\\6Fe^{2+}\rightarrow 6e^-+6Fe^{3+}$

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$Cr_2O_7^{-2}+6Fe^{2+}\rightarrow 2Cr^{3+}+6Fe^{3+}$

Now you must deal with the unbalanced O₇ on the left.  If you choose an acid medium you put H₂O on the right and H⁺ on the left, like this:

$Cr_2O_7^{-2}+6Fe^{2+}+14H^+\rightarrow 2Cr^{3+}+6Fe^{3+}+7H_2O$

If you choose a basic medium you put H₂O on the left and OH^- on the right, like this:

$Cr_2O_7^{-2}+6Fe^{2+}+7H_2O\rightarrow 2Cr^{3+}+6Fe^{3+}+14OH^-$