In an electrolytic cell (not galvanic/Voltaic), electric current is supplied fro

Question

In an electrolytic cell (not galvanic/Voltaic), electric current is supplied from an external source to force a non-spontaneous reaction to occur. On its own, saltwater does not spontaneously generate any solids or gases. However, on supplying current at the correct voltage, electrolysis can occur in an aqueous solution of sodium chloride. Na^+ + e^- rightarrow E_red degree = -2.76 V Cl_2 + 2e^- rightarrow 2Cl^- E_red degree = 1.36 V 2 H_2 O + 2 e^- rightarrow H_2 + 2 OH^- E_red degree = -0.83 V O_2 + 4H^+ + 4e^- rightarrow 2 H_2 O E_red degree = 1.23 V Which species present is most easily oxidized: Na^+, Cl^+, or H_2O? Which is most easily reduced? What products do you expect to observe, then? How much voltage must you apply to cause this reaction?

Explanation / Answer

Which species present is most easily oxidized Na+,cl- or h2o

Answer : Na+-------- oxidized.

The species with the lowest E value is the hardest to reduce .

its reverse reaction will have the species

which is the most easily oxidized.

Answer : H2O-------- reduced

The species with the higher E value is the most easily reduced

Molten NaCl, i.e. NaCl (l)

Ions present: Na+ and Cl–

Reduction at cathode: Na+(aq) + e– Na(s)

Oxidation at anode: 2Cl– Cl2 (g) + 2e–

Aqueous NaCl,

Ions present Na+ and Cl–

At the anode possible reactions are:

2Cl– (aq) Cl2 (g) + 2e– –E = –1.36 V

2H2 O(l) 4H+(aq) + O2 (g) + 4e– –E = –1.23 V

Based on the –E values,

the oxidation of H2 O will be preferred as it has the higher –E value.

At the cathode possible reduction reactions are:

Na+(aq) + e– Na(s) –E = –2.71 V

2H2 O(l) + 2e– H2 (g) + 2OH– (aq) –E = –0.83 V

Based on the E values, the reduction of H2 O will be preferred as it has the higher E value and H2 (g) will be discharged at the cathode.

At any concentration of NaCl(aq) the reaction at the cathode for aqueous NaCl is different to the reaction that occurs at the cathode with molten NaCl.

H2 O is preferentially reduced in the aqueous solution to form

H2 (g): 2H2 O(l) + 2e– H2 (g) + 2OH– (aq).

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