<p>I need to find the net ionic equation for the following balanced redox equation</p>
<p>2Al(s) + 6HI(aq) -> 2AlI3(aq) + 3H2(g)</p>
<p>I have precisely one try left on capa....:(....or maybe someone can tell me the correct way to enter charge/charge numbers on capa</p>
<p>thanks in advance</p>
<p>To start, I will right out the complete ionic equation on the reactants side. HI(aq) is a strong acid, and thus it dissociates into its ions.</p>
<p>2Al(s) + 6H+ + 6I- --></p>
<p>In the products, you know that H2 gas is produced, but what about Al and I-? According to the original equation, Al is oxidized to Al3+ when H+ is reduced, and I- remains I-.</p>
<p>--> 2Al3+ + 6I-** + 3H2
(**: there were two moles of AlI3, and 3I's, thus the coefficient is 6)</p>
<p>This is the result (the complete ionic equation):
2Al(s) + 6H+ + 6I- --> 2Al3+ + 6I- + 3H2</p>
<p>Now, cancel out the stuff that appears on both sides.</p>
<p>You're left with</p>
<p>2Al(s) + 6H+ --> 2Al3+ + 3H2</p>
<p>Which is the answer; i.e. the net ionic equation.</p>
<p>You should always check that the charges and the mass and the atoms are conserved throughout the equation, which indeed it is.</p>
<p>thanks for the help, I had trouble with the charge - +3 (oxidation number) instead of 3+ (ion charge)</p>
<p>got it right on the last try so it's all good :p</p>