After the chemistry lecture on single replacement redox reactions, your classmat
ID: 930588 • Letter: A
Question
After the chemistry lecture on single replacement redox reactions, your classmate tells you he has come up with a plan to produce gold and get rich. His plan involves ordering 325.00 mL of a solution of aqueous Au(NO3)3 {gold(lll) nitrate} at 0.0255 M and 13.457 g of Sn (tin) from a chemical company.
(a) Give the balanced molecular, ionic, and net ionic equations for a possible reaction between those two reactants including their physical states if tin has +2 charge in the product
(b) Show the balancing of the redox reaction (shown by the equation above) by separating the oxidation and reduction steps; use the net ionic equation.
(c) identify the oxidating agent and the reducing agent.
(d) Writing the molecular equation and using an arrow show the transfer of electrons from one reactant to another
(e) Which of the chemicals will be the 'limiting reactant'? Show all calculations clearly to support your answer.
(f) Which of the chemicals will be the 'excess reactant'? How much of the excess reactant will remain unreacted? Show calculations to support your answer.
(g) Calculate the maximum mass of gold that can produced from the supplies he will order. Show work clearly.
(h)The supplies arrive and your friend prepares to run the experiment. However, in the process of getting the lab set up, he gets interrupted and leaves the cap of the Au(NO3)3 solution by accident. He is unable to returns to the lab for another 2 weeks. in that time, 40.00mL of water evaporated and the aqueous Au(NO3)3 solution is now 285.00mL. If he runs the reaction now, would your answer to part (g) change? If so, by how much? if not, why not? Show calculations to support your answer in coherent complete sentences
Explanation / Answer
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) ½I2(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)