Question

Balance the following equations:
$Fe(CN{)}_6^{-4}+Mn{O}_4^{-}\to F{e}^{-3}+C{O}_2+N{O}_3^{-}+M{n}^{+2}$

Answer 1

$5\left[Fe\right(CN{)}_6{]}^{4-}+Mn{O}_4^{-}+8H^{+}\to 5\left[Fe\right(CN{)}_6{]}^{3-}+M{n}^{2+}+4H_{2}O$

Step 1. Write down the unbalanced equation ('skeleton equation') of the chemical reaction. All reactants and products must be known. For a better result write the reaction in ionic form.

Step 2. Separate the process into half reactions. A redox reaction is nothing but both oxidation and reduction reactions taking place simultaneously.

Step 3. Balance the atoms in each half reaction. A chemical equation must have the same number of atoms of each element on both sides of the equation. Add appropriate coefficients (stoichiometric coefficients) in front of the chemical formulas to balance the number of atoms. Never change any formulas.

Step 4. Make electron gain equivalent to electron lost. The electrons lost in the oxidation half-reaction must be equal the electrons gained in the reduction half-reaction. To make the two equal, multiply the coefficients of all species by integers producing the lowest common multiple between the half-reactions.

Step 5. Add the half-reactions together. The two half-reactions can be combined just like two algebraic equations, with the arrow serving as the equals sign. Recombine the two half-reactions by adding all the reactants together on one side and all of the products together on the other side.

Step 6. Simplify the equation. The same species on opposite sides of the arrow can be canceled. Write the equation so that the coefficients are the smallest set of integers possible.

Finally, always check to see that the equation is balanced. First, verify that the equation contains the same type and number of atoms on both sides of the equation.