# Au + I2+ → __________ Write reduction half reaction and the oxidation half reaction. Write reduction half reaction and the oxidation half reaction. Write the balanced redox equation. Calculate...

Au + I2+ → __________

• Write reduction half reaction and the oxidation half reaction.
1. Write reduction half reaction and the oxidation half reaction.
2. Write the balanced redox equation.
3. Calculate the cell potential, E°. Show your work.
4. Indicate whether the reaction is spontaneous or nonspontaneous.
Au + I2+ → __________

bandmanjoe | Certified Educator

When dealing with oxidation-reduction equations, it is helpful to remember the mnemonic "OIL RIG", which stands for "oxidation is loss, reduction is gain".  The element that loses or donates electrons is being oxidized, while the element gaining electrons is being reduced.  So the oxidation half equation would be this:  Au ---> Au+1  + 1e-, while the reduction half equation would be  2I + 2e- ---> 2I-- .  The balanced redox equation would look like this:

2Au+  + 2e-  + I2 --->  2AuI

The bond enthapy for gold is 225 kj mol-1, while iodine is 151 kj mol-1.  To calculate the cell potential, we add the energy needed to break the bonds of the reactants, then add the bonds of formation for the products:

151 x 2 to break apart the I2 equals +302.

151 + 225 = 376 kj mol-1 to form one molecule of AuI; times 2 molecules formed would be 752 kj mol-1.

Subtract the +302 from -752 and you get -450 kj mol-1 given off from the formation of AuI.

Since the energy given off is greater than the energy put into the reaction the reaction is exothermic and may be considered spontaneous.