Look the standard reduction potentials of each of the following half reactions f

ID: 500021 • Letter: L

Question

Look the standard reduction potentials of each of the following half reactions from the table in the back of your book (or my lecture slides) and write them here and also in your notebook. Make sure you pay attention to the charges, and choose the correct reactions from the table! Mg^2+ (aq) + 2 e^- rightarrow Mg_(s) E^0 red = Zn^2+ (aq) + 2 e^- rightarrow Zn_(s) E^0 red = Al^3+ (aq) + 3 e^- rightarrow Al_(s) E^0 red = Ag^+ (aq) + e^- rightarrow Ag_(s) E^0 red = Cu^2+ (aq) + 2 e^- rightarrow Cu_(s) E^0 red = Fe^3+ (aq) + 3 e^- rightarrow Fe_(s) E^0 red = Ni^2+ (aq) + 2 e^- rightarrow Ni_(s) E^0 red = You have two electrodes in an electrochemical cell. They are not yet connected by an external wire. You touch the red lead of the multimeter to "electrode A" and the black lead to "electrode B." The multimeter reads: "+ 0.34 V." Which electrode was the anode and which was the cathode? (This is covered in this lab write-up, not in class.) Write a balanced reaction with copper acting as an anode and silver acting as a cathode.

Explanation / Answer

3) The voltmeter explicitly measures the overall redox potential between the two half-cells. The voltmeter has two probes, a red and black probe and records the voltage difference between the red probe and the black probe. By convention, the potential at the black probe is assigned a voltage of 0.00 V. Therefore, the voltage measured is the potential of the red probe relative to 0.00 V.

If the measured voltage is 0.34 V, then we can assume that B electrode has 0.00 V and A has 0.34 V. From the above table we find that the E0Cu2+/Cu is 0.34 V while that of E0Fe3+/Fe is -0.04 V. Cu2+/Cu is at higher potential and hence is cathode while Fe3+/Fe is anode.

Ecell = Ecathode- Eanode

= 0.34 – (0.04) = 0.30 V

The value of Ecell is in close agreement with the measured voltage.

4) 2Ag+(aq) + 2e– ------> 2Ag(s)

Cu(s) ------> Cu2+ (aq) + 2e–

Overall reaction

2Ag+(aq) + Cu(s) ------> Cu2+(aq) + 2Ag(s)

Half reaction

E0red (V)

Mg2+(aq) + 2 e– ------>Mg(s)

-2.37

Zn2+(aq) + 2 e– ------>Zn(s)

-0.763

Al3+(aq) + 3 e– ------>Al(s)

-1.66

Ag+(aq) + e– ------>Ag(s)

+0.7994

Cu2+(aq) + 2 e– ------> Cu(s)

+0.337

Fe3+(aq) + 3 e– ------>Fe(s)

-0.04

Ni2+(aq) + 2 e– ------> Ni(s)

-0.25