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.
In order for us to help we need more information. Possibly the amount of grams you used for Mg in trial 1 or something similar to this.
As the atomic mass of iron is 55.847u
We say that it is the mass of one mole of iron.
By formula it can be find by
No.of mole=mass in g/molar mass
Mass in gram = no.of mole x molar mass
No.of mole =1
Molar mass = 55.847g
Mass in gram = 1x55.847
= 55.847g of Fe
