In the cell Pt(s)∣ Fe²⁺(aq)∣ Fe³⁺(aq) Cu²⁺(aq)∣ Cu(s)which will increase the cell voltage the most?

Question

Quizplus · Accepted Answer

The cell potential is determined by the difference in electrode potentials for the oxidation and reduction half-reactions. In this case, the two half-reactions are:
Pt(s) + 2 FeS1U1P12+(aq) + 2 e- → 2 FeS1U1P13+(aq)          (oxidation at anode)
CuS1U1P12+(aq) + e- → Cu(s) + S1S1P0(aq)                            (reduction at cathode)

To increase the cell potential, we need to increase the difference in electrode potentials. We can do this by either increasing the reduction potential (for the reduction half-reaction) or decreasing the oxidation potential (for the oxidation half-reaction).

Of the options given, doubling the concentration of FeS1U1P12+S1S1P0 will have the greatest effect on the oxidation half-reaction, as it will increase the concentration of the oxidizing agent (FeS1U1P12+). This will shift the equilibrium towards the reduction half-reaction, increasing the cell potential.

Therefore, the best choice is D (double [FeS1U1P12+S1S1P0]).

In the Cell Pt(s)∣ Fe2+(aq)∣ Fe3+(aq) Cu2+(aq)∣ Cu(s)which Will

In the Cell Pt(s)∣ Fe²⁺(aq)∣ Fe³⁺(aq) Cu²⁺(aq)∣ Cu(s)which Will