Print Calculator Periodic Table X Incorrect Incorrect Question 16 of 36 Map sapl

Question

Print Calculator Periodic Table X Incorrect Incorrect Question 16 of 36 Map sapling learning Complete the half reactions for the cell shown here, and show the shorthand notation for the cell by dragging labels to the correct position. The electrode on the left is the anode, and the one on the right is the cathode Anode half-reaction: Pb 2e Cathode half-reaction: 2C17 AgCl(s) PbCl2(s KCI (aq) KCl(aq) Shorthand notation: 2AgCl (s) Pb(s) Previous Give Up & View Solution Check Answer Next Exit Hint

Explanation / Answer

Since the oxidation takes place in the anode and the reduction in the cathode.

The reaction in the anode would be Pb(s) --> Pb2+ + 2e- (oxidation)

The reaction in the cathode would be Ag+ + e- --> Ag (reduction)

The shortand notation is always from the next way   Anode reaction II cathode reaction

For your problem it would be: Pb(2) | Pb(2+)|KCl || KCl | Ag+ | Ag

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