In the voltaic cell that is represented as: Cu | Cu2+ || Fe | Fe2+.

The only anion in solution in the cells and salt bridge is the nitrate ion.
The directions the electron external circuit and NO3- flow in the salt bridge circuit will be:

*electron flow is Fe to Cu
*electron flow is Cu to Fe
*electron flow is cathode to anode.
*electron flow is anode to cathode.
*ion flow is NO3- ions to the cell with the anode
*ion flow is NO3- ions to the cell with the cathode

- The electron flow in the external circuit is from the anode to the cathode. The anode is the Fe electrode, and the cathode is the Cu electrode. Therefore, the electron flow is from Fe to Cu.

- The ion flow in the salt bridge circuit is driven by the movement of the nitrate (NO3-) ions. The nitrate ions will move from the cell containing the anode (Fe) to the cell containing the cathode (Cu). Therefore, the ion flow is NO3- ions to the cell with the cathode.

To determine the direction of electron flow in the voltaic cell, you need to look at the oxidation and reduction half-reactions occurring at each electrode.

In this voltaic cell, the half-reaction occurring at the cathode (Cu electrode) is the reduction of Cu2+ ions to form Cu metal:

Cu2+ + 2e- -> Cu

The half-reaction occurring at the anode (Fe electrode) is the oxidation of Fe atoms to form Fe2+ ions:

Fe -> Fe2+ + 2e-

Now, based on the standard conventions for voltaic cells:

- Electrons always flow from the anode (oxidation) to the cathode (reduction).
- The direction of electron flow is opposite to the direction of the ion flow in the salt bridge.

Therefore, since the reduction half-reaction is occurring at the Cu cathode and the oxidation half-reaction is occurring at the Fe anode, the correct statement is:

* Electron flow is anode to cathode.

Regarding the ion flow in the salt bridge, nitrate ions (NO3-) are the only anions present. The ion flow in the salt bridge is driven by the need to maintain electrical neutrality in the half-cells. Positively charged ions flow towards the half-cell with a higher concentration of negative charge to balance the charge. Similarly, negatively charged ions flow towards the half-cell with a higher concentration of positive charge to maintain electrical neutrality.

Since Fe2+ ions are being produced at the Fe anode (which is the half-cell with a higher concentration of positive charge), the correct statement is:

* Ion flow is NO3- ions to the cell with the anode.

The rxn is

Fe + Cu^2+ ==> Cu + Fe^2+
so the electrons flow from Fe to Cu^2+. The anode is where oxidation occurs.
Since the Fe is going into solution and the Cu^2+ is coming out of solution, the Cu side has too many nitrate ions.