The first step in balancing any redox reaction is determining whether or not it is even an oxidation-reduction reaction, which requires that species exhibits changing oxidation statesduring the reaction. To maintain charge neutrality in the sample, the redox reaction will entail both a reduction component and an oxidation components and is often separated into independent two hypothetical <span>half-reactions </span>to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:
<span><span><span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span>(1)</span><span>(1)<span>Cu(s)+2A<span>g+</span>(aq)→C<span>u<span>2+</span></span>(aq)+2Ag(s)</span></span></span>
The first step in determining whether the reaction is a redox reaction is to splitting the equation into two hypothetical half-reactions. Let's start with the half-reaction involving the copper atoms:
<span><span><span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span>(2a)</span><span>(2a)<span>Cu(s)→C<span>u<span>2+</span></span>(aq)</span></span></span>
The oxidation state of copper on the left side is 0 because it is an element on its own. The oxidation state of copper on the right hand side of the equation is +2. The copper in this half-reaction is oxidized as the oxidation states increases from 0 in Cu to +2 in Cu2+. Now consider the silver atoms
<span><span><span>2A<span>g+</span>(aq)→2Ag(s)</span>(2b)</span><span>(2b)<span>2A<span>g+</span>(aq)→2Ag(s)</span></span></span>
In this half-reaction, the oxidation state of silver on the left side is a +1. The oxidation state of silver on the right is 0 because it is an element on its own. Because the oxidation state of silver decreases from +1 to 0, this is the reduction half-reaction.
Consequently, this reaction is a redox reaction as both reduction and oxidation half-reactions occur (via the transfer of electrons, that are not explicitly shown in equations 2). Once confirmed, it often necessary to balance the reaction (the reaction in equation 1 is balanced already though), which can be accomplished in two ways because the reaction could take place in neutral, acidic or basic conditions.