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Ok this is about electrolysis of sodium Sulphate
I believe the formula is
Na2SO4 + 2H2O ----> 2NaOH + H2SO4
During the lab, the indicator at the cathode turned basic and the anode indicated acidic.
Explaining this using electric potentials:
Na + e = Na E = -2.71 V
H2O + e = ½ H2 + OH- E = -0.83 V
S2O8 + 2e = 2SO4 E = 2.05 V
½ O2 + 2H = H2O E = 1.23 V
water = 1.23 – (-0.83) = 2.06 V
sodiumSulphate = 2.05 – (-2.71) = 4.76 V
The battery used were 3V
I have no idea how this results in the original reaction formula above. Can someone englighten me?
I believe the formula is
Na2SO4 + 2H2O ----> 2NaOH + H2SO4
During the lab, the indicator at the cathode turned basic and the anode indicated acidic.
Explaining this using electric potentials:
Na + e = Na E = -2.71 V
H2O + e = ½ H2 + OH- E = -0.83 V
S2O8 + 2e = 2SO4 E = 2.05 V
½ O2 + 2H = H2O E = 1.23 V
water = 1.23 – (-0.83) = 2.06 V
sodiumSulphate = 2.05 – (-2.71) = 4.76 V
The battery used were 3V
I have no idea how this results in the original reaction formula above. Can someone englighten me?
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