103A Results: Part B (Concentration Effects) Experiment 14: Electrochemical Cell
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Question
103A Results: Part B (Concentration Effects) Experiment 14: Electrochemical Cells Show the diagrams for your cell. a) the chemicals b) the direction of electron flow c) the anode and cathode d) the e) the half reactions that occur in each half-cell 0 the overall cell reaction movement of the anions and cations through the porous barrier w the cell diagram for the solution I cell (the copper ion dilution) below, draw the cell diagram for solution 2 in your lab notebook. o.1a3 Cathode Anode ) ations 0.5 M Ha S04 Part B: Data/Calculated values Solution 2 and 0.5M Ca Cells: Solution 1 and 0.5M Cu Ecell observed) (calculated) Question: Discuss the effect of a concentration gradient on the E what happens when you make larger concentration gradient? Explain why a concentration gradient causes electron flowExplanation / Answer
The positive electrode is SHE or standard hydrogen electrode,with Pt electroode in 0.5MH2SO4.
Cell reaction for this electrode:
2H+(aq)+2e ---->H2(g) Eo(red)=0.00V
The negative electrode is the Cu electrode with CuSO4 as the electrolyte.Cell reaction:
Cu2+(aq) +2e --->Cu(s), Eo(red)=+0.34V
As the reduction potential of Cu2+|Cu electrode > the reduction potential of SHE so Cu2+|Cu will be the cathode (reduction is favored) and SHE will be the anode.
Cu2+(aq) +2e --->Cu(s), Eo(red)=+0.34V [cathode]
H2(g)---> 2H+(aq)+2e Eo(ox)=0.00V [anode]
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net cell rxn: Cu2+(aq)+ H2(g) ---->Cu(s)+2H+(aq) Ecell=E(red)+(Eox)=0.34+0.00=0.34V
Ecell=0.34V [Ecell calculated]
As externally electrons are coming out of or released from (H2(g)---> 2H+(aq)+2e ) Also the formation of H+ ions attracts anions from the salt bridgein H2|H+ half cell so it is negative .
Cu2+|Cu electrode is consuming electrode so it is the positive ,with the reduction of cations ,so SO42- anions are remaining in the solution, which attracts positive ions from the salt bridge.(positive electrode)
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