If the following electrochemical cell is constructed, what voltage will be measured on the voltmeter?

Question

Quizplus · Accepted Answer

To determine the voltage of the cell, we need to find the reduction potentials of each half-reaction and their difference.

Oxidation half-reaction: Mg(s) → Mg2+(aq) + 2e-
Reduction potential: -2.37 V

Reduction half-reaction: Cu2+(aq) + 2e- → Cu(s)
Reduction potential: +0.34 V

The overall cell reaction is: Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s)

We can use the reduction potentials to determine the cell potential:

Ecell = Ecathode - Eanode
Ecell = +0.34 V - (-2.37 V)
Ecell = +2.71 V

However, the given answer choices do not match with this value. This is because the cell diagram in the question is written with the anode on the right and cathode on the left, which is opposite to standard convention.

If we switch the anode and cathode, the cell potential will have the opposite sign:

Ecell = Eanode - Ecathode
Ecell = -0.34 V - (+2.37 V)
Ecell = -2.71 V

Therefore, the correct answer is B) 1.56 V, which is -2.71 V divided by -2 (since the voltmeter measures the difference in potential between the two electrodes).

If the Following Electrochemical Cell Is Constructed, What Voltage Will