Complete and balance the following equation:
Cr2O7{2-}(aq)+CH3OH(aq)-->HCO2H(aq)+Cr{3+}(aq)
Any feedback would be great, thank you so much.
Here is everything about redox including oxidation numbers and how to balance half equations.
To balance the equation, we need to ensure that the number of atoms of each element is the same on both sides of the equation. Let's go through the process step by step:
First, let's balance the carbon (C) atoms. There is one carbon atom on the left side from CH3OH, so we need one carbon atom on the right side. We can achieve this by placing a coefficient of 1 in front of HCO2H:
Cr2O7{2-}(aq) + CH3OH(aq) --> HCO2H(aq) + Cr{3+}(aq)
Next, let's balance the hydrogen (H) atoms. On the left side, there are four hydrogen atoms from CH3OH, so we need four hydrogen atoms on the right side. We can achieve this by placing a coefficient of 4 in front of HCO2H:
Cr2O7{2-}(aq) + CH3OH(aq) --> 4HCO2H(aq) + Cr{3+}(aq)
Now, let's balance the oxygen (O) atoms. On the left side, there are seven oxygen atoms from Cr2O7{2-}, and one oxygen atom from CH3OH, making a total of eight oxygen atoms. On the right side, there are four oxygen atoms from 4HCO2H. So, we need to balance the oxygen atoms by adding four more to the left side. We can achieve this by placing a coefficient of 4 in front of Cr2O7{2-}:
4Cr2O7{2-}(aq) + CH3OH(aq) --> 4HCO2H(aq) + Cr{3+}(aq)
Finally, let's balance the chromium (Cr) atoms. On the left side, we have four chromium atoms from 4Cr2O7{2-}, so we need four chromium atoms on the right side. Therefore, we can balance the equation by placing a coefficient of 4 in front of Cr{3+}:
4Cr2O7{2-}(aq) + CH3OH(aq) --> 4HCO2H(aq) + 4Cr{3+}(aq)
This is the balanced equation for the reaction.