Anode:
"2H_2O(l) \\rightarrow O_2(g) + 4H^+(aq) + 4e^-"Cathode:
"2 \\times (2H^+ + 2e^- \\rightarrow H_2)"Calculate the moles of electrons when a current of 5A is passed through acidified water for 198seconds:
Calculate the moles of O2:
"0.0103 mol e^- (\\frac{1 mol O_2}{4 mol e^-}) = 0.00256 mol"Calculate vollume of O2:
"V = V_m \\times n = 22.4 \\frac{dm^3}{mol}\\times 0.00256 mol = 0.0575 dm^3"Answer: "0.0575 dm^3"
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