How to balance equations for reduction/oxidation reaction?

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 half-reactions to aid in understanding the reaction. This requires identifying which element is oxidized and which element is reduced. For example, consider this reaction:

Cu (s) + 2A g + (aq) →C u 2 + (aq) + 2Ag (s) (1) (1) Cu (s) + 2A g + (aq) →C u 2 + (aq) + 2Ag (s)

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:

Cu (s) →C u 2 + (aq) (2a) (2a) Cu (s) →C u 2 + (aq)

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

2A g + (aq) →2Ag (s) (2b) (2b) 2A g + (aq) →2Ag (s)

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.