The following galvanic cell: Zn \| Zn(NO3)2 (100 ml, 1 M) \|\| Cu(NO3)2 (100 ml, 1 M) \| Cu was oper — Electrochemistry Chemistry Question
Question
The following galvanic cell: Zn \| Zn(NO3)2 (100 ml, 1 M) \|\| Cu(NO3)2 (100 ml, 1 M) \| Cu was operated as an electrolytic cell as Cu as the anode and Zn as the cathode. A current of 0.4825 A was passed for 10 h and then the cell was allowed to function as galvanic cell. What would be the final EMF of the cell at 25°C? Assume that the only electrode reactions occurring were those involving Cu \| Cu^2+ and Zn \| Zn^2+. Given: E°_Cu2+\|Cu = +0.34 V, E°_Zn2+\|Zn = –0.76 V, 2.303RT/F = 0.06, log 1.9 = 0.28)
💡 Solution & Explanation
During electrolysis, Cu oxidizes to Cu^2+ and Zn^2+ reduces to Zn. Total charge Q = 0.4825 A × 10 × 3600 s = 17370 C. Faradays = 17370 / 96500 = 0.18 F. Moles of Cu^2+ produced = 0.18/2 = 0.09 mol. New Cu^2+ moles = 0.1 + 0.09 = 0.19 mol (concentration = 1.9 M). Moles of Zn^2+ consumed = 0.09 mol. New Zn^2+ moles = 0.1 - 0.09 = 0.01 mol (concentration = 0.1 M). Final galvanic cell EMF = E° - (0.06/2)*log([Zn^2+]/[Cu^2+]) = 1.10 - 0.03*log(0.1/1.9) = 1.10 + 0.03*log(19) = 1.10 + 0.03*(log 1.9 + 1) = 1.10 + 0.03*(0.28 + 1) = 1.10 + 0.0384 = 1.1384 V. Therefore, correct answer is C.