The oxidation of gold metal by aqua regia (i.e., royal water) results in the pro

Question

The oxidation of gold metal by aqua regia (i.e., royal water) results in the production of nitrogen dioxide and gold(III) iron, in the primary net ionic step. Aqua regia is a dangerous mixture of 1 part nitric acid and 3-9 parts hydrochloric acid. Ignore the activity of the chloride ion. Is the nitrate ion a spectator ion or a participating ion, or both? A. Print the oxidation half-equation B. Print the reduction half-equation C. Print the net ionic equation (without any spectator ions unless they are involved in the balance Please include explanation!

Explanation / Answer

It is not always possible to balance redox equations using the simple inspection technique. The following unbalanced net ionic equation provides an example. Au3+(aq) + I-(aq) --> Au(s) + I2(s) At first glance, it seems that this equation can be balanced by placing a 2 in front of the I-. Au3+(aq) + 2I-(aq) --> Au(s) + I2(s) Note, however, that although the atoms are now balanced, the charge is not. The sum of the charges on the left is +1, and the sum of the charges on the right is zero, as if the products could somehow have one more electron than the reactants. To correctly balance this equation, it helps to look more closely at the oxidation and reduction that occur in the reaction. The iodine atoms are changing their oxidation number from -1 to 0, so each iodide ion must be losing one electron. The Au3+ is changing to Au, so each gold(III) cation must be gaining three electrons. The half-reactions are: I-(aq) --> I2(s) + e- Au3+(aq) + 3e- --> Au(s) We know that in redox reactions, the number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidizing agent; thus, for each Au3+ that gains three electrons, there must be three I- ions that each lose one electron. If we place a 3 in front of the I- and balance the iodine atoms with a 3/2 in front of the I2, both the atoms and the charge will be balanced. Au3+(aq) + 3I-(aq) --> Au(s) + 3/2I2(s) or 2Au3+(aq) + 6I-(aq) --> 2Au(s) + 3I2(s)

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