A voltaic cell is constructed that uses the following half-cell reactions:
Cu⁺(aq) + e^- → Cu(s)
I₂(s) + 2 e^- → 2 I^-(aq)
The cell is operated at 298 K with [Cu⁺] = 0.25 M and [I^-] = 0.035 M.
(b) Which electrode is the anode of the cell?
(c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions?

Verified step by step guidance

1. Identify the oxidation and reduction half-reactions. In the given cell, Cu+(aq) + e- → Cu(s) is the reduction half-reaction, and I2(s) + 2 e- → 2 I-(aq) is the oxidation half-reaction.
2. Determine which electrode is the anode. The anode is where oxidation occurs. Since I2 is being reduced to I-, the I2/I- half-cell is the anode.
3. Consider the standard cell potential (E°) for each half-reaction. Under standard conditions, the half-reaction with the lower (more negative) standard reduction potential is the anode.
4. Compare the cell operation under non-standard conditions to standard conditions. Under non-standard conditions, the concentrations of the ions can affect the cell potential, but the identity of the anode and cathode is determined by the direction of electron flow.
5. Conclude whether the anode remains the same under standard conditions. Since the identity of the anode is based on the inherent tendency of the half-reactions to oxidize or reduce, it should remain the same under standard conditions.