What happens to iron electrodes (attached to a battery) in copper(II) sulphate solution? After a few hours, the following observations were made:


The solution turns green. My guess: $\ce{FeSO4}$ is formed.
There are reddish-brown and black powder deposits on both the electrodes.
The deposits fall down easily when the set-up is shaken a little.


What are the reactions that are occurring?

According to me, the copper is displaced from the $\ce{CuSO4}$ and is deposited on the iron electrode (the cathode), and $\ce{FeSO4}$ is formed as a green solution. What is getting deposited on the anode? Are there ferrous and ferric oxides/hydroxides being formed?
The reaction is spontaneous, you don't need put a power supply.

Let's see the reaction with the potentials (obtained from the table of standard reduction potentials):

$$
\begin{align}
\ce{Fe^0 &amp;→ Fe^2+ + 2 e-} &amp;\quad E^\circ &amp;= \pu{0.44 V} \\
\ce{Cu^2+ + 2 e- &amp;→ Cu^0} &amp;\quad E^\circ &amp;= \pu{0.337 V} \\
\hline
\ce{Fe^0 + Cu^2+ &amp;→ Fe^2+ + Cu^0} &amp;\quad E^\circ &amp;= \pu{0.777 V}
\end{align}
$$

So, the solid formed in both electrodes is copper. The green solution is related to iron sulfate.