Homework Help Forum: chemistry

Posted by sh on Wednesday, November 4, 2009 at 10:51pm.

You have 5 metals A, B, C, D, E in 1.00M solutions of A^(3+), B^(3+), C^(3+), D^(3+), and E^(3+) respectively. When some of the half-cells are connected, the following results are found. Cathode /Anode /Cell Voltage C E 0.10V D B 0.08V A C 1.41V D C 0.55V Arrange the ions in decreasign strength as oxidizing agents. How do you find the cell potential when they're all unknown?

• chemistry - DrBob222, Wednesday, November 4, 2009 at 11:12pm

  You must compare them. If C is the anode and A is the cathode, you have 1.41 volts generated; therefore, C is a better reducing agent than A (or A is a better oxidizing agent than C). So on a chart, place A --stronger oxidizing agent. C --weaker oxidizing agent. Then do the others and fill in the gaps.

  
  

• chemistry - sh, Wednesday, November 4, 2009 at 11:22pm

  After comparing them I came up with A^(3+)>D^(3+)> ? > ? > E^(3+) I can't figure out whether B is the stronger oxidizing agent or C.

  
  

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