Question #0853f

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Question #0853f

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Answer 1 · 2015-07-20T18:06:51

You can use it like this:

Explanation:

Equation:

$2 A l_{(s)} + 6 H C l_{(a q)} \to 2 A l C l_{3 (a q)} + 3 H_{2 (g)}$ $2Al_((s))+6HCl_((aq))rarr2AlCl_(3(aq))+3H_(2(g))$

Aluminium loses it's outer electrons:

$2 A l_{(s)} \to 2 A l_{(a q)}^{3 +} + 6 e$ $2Al_((s))rarr2Al_((aq))^(3+)+6e$

These are taken in by the hydrogen ions:

$6 H_{(a q)}^{+} + 6 e \to 3 H_{2 (g)}$ $6H_((aq))^++6erarr3H_(2(g))$

In this reaction the chloride ions are spectators - they take no part in the reaction.

Answer 2 · 2015-07-20T19:07:43

If you want to be able to balance the equation using oxidation states, here's how...

Explanation:

First you need to work out the oxidation states of each element, notice the ones that have changed, work out how many electrons they have lost/gained. From there find the key number to balance the equation and from then on this trial and error.

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Question #0853f

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