Write balanced half-cell reactions and equations for the following:
Mo24O37(s) + MnO4-(aq) makes MoO3(s) + Mn+2(aq) (BASIC)
whats the answer
i figured it out
9Mo24O37+68H^++68MnO4^----->9Mo24O3+34H2o+68Mn^2+
To write balanced half-cell reactions and equations, we need to first understand the different species present and their oxidation numbers. Let's start with the given reaction:
Mo24O37(s) + MnO4-(aq) → MoO3(s) + Mn+2(aq) (BASIC)
From the reaction, we can see that the reduction of MnO4-(aq) to Mn+2(aq) takes place. The reduction half-reaction involves the gain of electrons.
We can identify the following species and their respective oxidation numbers:
- MnO4-(aq): Mn = +7
- Mn+2(aq): Mn = +2
To balance the half-reaction, we need to equalize the number of charges on both sides by adding electrons (e-):
MnO4-(aq) + 5e- → Mn+2(aq)
Now, let's move on to the oxidation half-reaction, which involves the oxidation of Mo24O37(s) to MoO3(s). The oxidation half-reaction involves the loss of electrons.
We can identify the following species and their respective oxidation numbers:
- Mo24O37(s): Mo = +6 (average oxidation state)
- MoO3(s): Mo = +6
To balance the half-reaction, we need to equalize the number of charges on both sides by adding electrons (e-):
Mo24O37(s) + 31e- → MoO3(s)
Finally, let's combine both half-reactions and ensure that the number of electrons lost equals the number of electrons gained:
3MnO4-(aq) + 31e- + 31Mo24O37(s) + 31e- → 31MoO3(s) + 3Mn+2(aq)
Now, let's simplify the equation by canceling out the electrons on both sides:
3MnO4-(aq) + 31Mo24O37(s) → 31MoO3(s) + 3Mn+2(aq)
This is the balanced half-cell reaction and equation for the given reaction in a basic solution.