Balance the ewuation using thr oxidation number method
Here is what i did
Au+ + H20---> Au2O3 + Au + H+
Au+ losses 3e- to form Au2O3-->oxidation
Au+ gets reduced to for Au
therefore the balanced chemical equation is 3Au+ + 3H20---> Au2O3 +Au +6H+
Does it make sense?
Yes and no.
The equation isn't balanced. (I know it looks balanced at first glance but it isn't.)
To be balanced, an equation must
1. have the same number of atoms on each side
2. have the same charge on each side.
3. electrons lost = electrons gained.
Your equation has 1 ok.
The charge doesn't balance (3+ on left and 6+ on right).
Electron change doesn't balance.
Here is the way I would do it.
Au^+ ==> Au2O3
First you make the Au atoms equal so we don't "split" charges.
2Au^+ ==> Au2O3
Oxidation state is +2 on left for 2 atoms and +6 on right for 2 atoms. We balance the change in state with electrons.
2Au^+ ==> Au2O3 + 4e
Count the charge on the left (+2) and right (-4) and balance with H^+.
2Au^+ ==> Au2O3 + 4e + 6H^+
Now add water to the other side to balance the H^+ added.
3H2O + 2Au^+ ==> Au2O3 + 4e + 6H^+
Next half cell.
Au^+ + e ==> Au
Now multiply the first equation by 1 and the second equation by 4 and add to obtain
3H2O + 6Au^+ ==> Au2O3 + 4Au + 6H^+
1. atoms balance 6H left and right. 3 O left and right. 6 Au left and right.
2. charge balances. +6 on left and right.
3.electron change balances. 2Au^+ on left move to Au2O3 where both Au atoms are +6. Loss of 4e. Other 4 Au^+ atoms move to 4Au on right. Gain of 4e.
Done.
Yes, your approach to balancing the equation using the oxidation number method makes sense. Let's go through the steps to ensure the equation is balanced correctly.
Step 1: Assign oxidation numbers to each element in the reaction. Since the oxidation number of oxygen is usually -2, and hydrogen is usually +1, we can assign these values first. Therefore, we have:
Au+ + H2O → Au2O3 + Au + H+
Step 2: Identify the elements whose oxidation numbers change. In this case, Gold (Au) undergoes both oxidation and reduction. Initially, Au has an oxidation state of +1, and in the product Au2O3, its oxidation state is +3. Hence, it loses 2 electrons, so the oxidation half-reaction is:
Au+ → Au3+ + 3e-
On the other hand, Au gets reduced from Au+ to Au in the product, so the reduction half-reaction is:
Au+ + e- → Au
Step 3: Balance the number of electrons lost and gained in each half-reaction. To equalize the number of electrons, we can multiply the oxidation half-reaction by 1, and the reduction half-reaction by 3:
3Au+ → 3Au3+ + 9e-
3Au+ + 3e- → 3Au
Since we need 9 electrons in the reduction half-reaction to match the 9 electrons in the oxidation half-reaction, we can now add the two half-reactions together:
3Au+ + 3e- + H2O → 3Au3+ + 9e- + H+
3Au+ + H2O → 3Au3+ + H+
Step 4: Balance the remaining elements by adjusting coefficients. To balance hydrogen atoms, we need to add 6 H+ ions on the reactant side:
3Au+ + H2O → 3Au3+ + H+
Finally, balance oxygen atoms by adding 3 H2O molecules on the product side:
3Au+ + 3H2O → Au2O3 + 3Au + H+
And there you have it! The balanced chemical equation using the oxidation number method is:
3Au+ + 3H2O → Au2O3 + 3Au + H+