We’re being asked to sketch a voltaic cell for the following reaction:
2 NO3–(aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H2O(l) + 3 Cu2+(aq)
Recall that a voltaic or galvanic cell is an electrochemical cell that is spontaneous and produces electricity.
From the reaction, we can see that the oxidation state of Cu went from 0 (Cu) to +2 (Cu2+) which means it was oxidized or lost electrons.
The anode (–) half-reaction is:
Cu(s) → Cu2+(aq) + 2 e–
0 = +2 + x
x = –2 = 2 e– to the product side
The other half-reaction is then reduced or gained electrons.
The cathode (+) half-reaction is:
2 NO3–(aq) + 8 H+(aq) + 6 e– → 2 NO(g) + 4 H2O(l)
2(–1) + 8(+1) + x = 0
x = –6 = 6 e– to the reactant side
Sketch a voltaic cell for each redox reaction. Label the anode and cathode and indicate the half-reaction that occurs at each electrode and the species present in each solution. Also indicate the direction of electron flow.
c. 2 NO3– (aq) + 8 H+(aq) + 3 Cu(s) → 2 NO(g) + 4 H 2O (/) + 3 Cu2+(aq)
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Based on our data, we think this problem is relevant for Professor McClure's class at DREXEL.
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Our data indicates that this problem or a close variation was asked in Chemistry: A Molecular Approach - Tro 3rd Edition. You can also practice Chemistry: A Molecular Approach - Tro 3rd Edition practice problems.