First fill in your half cell and cell reactions. F in standard cell potentials a

Question

First fill in your half cell and cell reactions. F in standard cell potentials as you calculate them. Oxidation at the Anode: The black (.) lead is attached to the electrode, which is the source of electrons. Write the anode half reaction: Reduction at the Cathode: The red lead (+) is attached to the electrode. Write the cathode half reaction: E degree - V. Overall Cell Reaction (Net Ionic equation): Write the overall cell reaction (balance electrons and add together): E degree_ cell = V E_ cell is the value measured by the voltmeter. If you had a standard cell (all concentration equal to 1 M), then E degree_ Cell is the value from the voltmeter. If your cell was not a standard cell, calculate your value of Q, then use the Nernst equation to find E degree_ cell. Assuming a standard copper cell, Cu(s) | Cu^2+ (1 M), if the copper electrode is the anode (black, -), then E degree_ cell = -0.34 V. If copper is the cathode, then E degree_ red = +0.34 V. We know that E degree_ cell = E degree_ cell + E degree_ red, so find the standard potential of the other half cell. Finding Reduction Potential (if necessary) for non-copper electrode: By convention, we report standard reduction potentials for half cells. If your non-copper electrode is not already a reduction reaction, write the reduction reaction (reverse equation) and find the reduction potential (change sign):

Explanation / Answer

Oxidation:

Mg2+(aq) + 2e– Mg(s) –2.38

since it has lower potential

the actual reacition (oxidation)

Mg(s)Mg2+(aq) + 2e- E = +2.38 V

For reduction:

Cu2+(aq) + 2e– Cu(s) +0.34

since it has a higher potential

The overall Cell reaction:

Mg2+(aq) + Cu2+(aq) + 2e– Cu(s) + Mg2+(aq) + 2e

E° = Ered + Eox = 0.34 + 2.38 = 2.72 V

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