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In part A of Expt. 10, we were going to construct some simple galvanic cells. The...

  1. In part A of Expt. 10, we were going to construct some simple galvanic cells. The first cell to be constructed consisted of Cu and Pb electrodes placed in 0.10 M solutions of Cu(NO3)2 and Pb(NO3)2 . Which electrode do you expect to be the anode and which would be the cathode? Explain.
  2. Can a galvanic cell have a negative cell potential? What should you do if you measured a negative cell potential when the two half-cells in a galvanic cell are connected to the coloured electrode clips in the lab?
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Answer #1

Given electrodes are cu and Pb Electrolytes are Cu(NO3)2 & Pb(NO3)2 4 current Pb W(ROO) (Cathode) (rod) - - - - Canode) 12 cu.Ce act as Cathode pb act as anode Ecel = Cathode - Famode - 12 Euro e prope = 0.342 – (-0.126) = 0.342 +0.126 = +0.468 v Cal

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