Recall the thecathodeis where reduction occurswhile theanodeis where oxidation occurs.

You are watching: Current is applied to an aqueous solution of calcium sulfide.

Since Na2S is anionic compound, it formsions when dissociating in water.

Na2S(s)→2Na+(aq) + S2–(aq)

Duringelectrolysis, thecation is reduced/gains electronswhile theanion is oxidized/loses electrons.

This means:

Na+ is reduced:Na+(aq) + 1 e–→Na (s);E˚ = -2.71 V

S2–is oxidized:S(s) + 2 H+ (aq) + 2 e-→H2S (g) E˚ = 0.14 V

Oxidation: H2S (g) → S(s) + 2 H+ (aq) + 2 e-→E˚ = -0.14 V(change sign)S2-as H2Soxidation is reverse of reductionThe typical reduction potentials (E˚) deserve to be uncovered in textbooks or online
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Problem Details

Current is applied to one aqueous equipment of sodium sulfide.

What is produced at the cathode?

a) O2(g)

b) H2(g)

c) Na(s)

d) S(s)

What is produced at the anode?

a)H2(g)

b) S(s)

c) Na(s)

d) O2(g)


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