Question

How to balance equations for reduction/oxidation reaction?

Answer 1

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)+2Ag+(aq)→Cu2+(aq)+2Ag(s)(1)(1)Cu(s)+2Ag+(aq)→Cu2+(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)→Cu2+(aq)(2a)(2a)Cu(s)→Cu2+(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

2Ag+(aq)→2Ag(s)(2b)(2b)2Ag+(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.