Determining formula of potassium molybdate from the amount of metal deposited during electrolysis

$$\ce{K2MoO_n}$$ in an acidified solution is electrolysed, $$\pu{0.112 L}$$ of oxygen is liberated and $$\pu{0.32 g}$$ of $$\ce{Mo}$$ is deposited (Atomic mass = $$96$$). Find $$n.$$

Now for this, I calculated that $$5×10^{-3}$$ moles of $$\ce{O2}$$ was liberated and $$\frac{10^{-2}}{3}$$ moles of Mo and since

1mol Mo $$\to \frac{n}{2}$$mol$$\ce{O2}$$

$$\frac{10^{-2}}{3}$$ moles of Mo$$\to \frac{10^{-2}n}{6}$$mol$$\ce{O2} = 5×10^{-3}$$moles of $$\ce{O2}.$$

This gives $$n = 3,$$ but the answer given was $$4.$$ Please help, where have I gone wrong?

• I'm kind of skeptical that Mo can be formed here at all. Not only molybdate is an anion and is likely not going to easily travel to cathode, and even if it would (say, by adding a third-party chelating ligand), in solution hydrogen formation is likely to dominate anyway, so I'd assume the best one can get out of this is $\ce{Mo2O3}.$ And yeah, you probably want to take electrolysis of water into account when estimating for the source of $\ce{O2}.$ – andselisk Aug 17 at 17:11
• @andselisk I think its a question to test the understanding in a topic like electrochemistry, so maybe that's why whosoever made this question, didn't think too much about it and , and oH yeah , you are right I forgot to consider Water!, Thanks – RandomAspirant Aug 17 at 19:08