Consider the following electrolysis reactions Consider the following electrolysi

Question

Consider the following electrolysis reactions

Consider the following electrolysis reactions Cathode Cu^2+ (aq, 0.186 M) + 2e^- Equilibrium Cu(s) Anode H_2O(l) Equilibrium 1/2 O_2(g, 1 bar) + 2H^+ (aq, 0.191 M) + 2e^- whose standard reduction potentials can be found in this table. Calculate the voltage needed to drive the net reaction if the current is negligible. Number ________ V The cell has a resistance of 53.0 Ohm and a current of 8.27 mA when the cathode overpotential is 0.335 V and the anode overpotential is 0.055 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Number _______ V

Explanation / Answer

Q.1: E0(red, Cu2+/Cu) = +0.34 V

E0(oxi, H2O/O2,2H+) = - 1.23 V

=> E0(cell) =  E0(red, Cu2+/Cu) + E0(oxi, H2O/O2,2H+) = 0.34 - 1.23 = - 0.89 V

The overall cell reaction is

Cu2+(aq, 0.186M) + H2O(l) ------> Cu(s) + 1/2 O2(g) + 2H+(aq, 0.191M) :  E0(cell) = - 0.89 V

[Cu2+(aq)] = 0.186 M

[H+(aq)] = 0.191 M

PO2 = 1 bar

The electrode potential can be calculated from Nernst equation:

E(cell) = E0(cell) + (0.0591 / n) x log ([H+(aq)]2 x PO2 / [Cu2+(aq)])

=> E(cell) = - 0.89 V + (0.0591 / 2) x log[(0.191)2 x 1 / (0.186)]

=> E(cell) = - 0.911 V

Since current is negligible, voltage required to drive the net reaction is + 0.911 V (answer)

Q.2: Cell overpotential = Cathode overpotential + anode overpotential = 0.335 + 0.055 = 0.390 V

Voltage required to flow 8.27 mA current = IxR = 8.27 x 10-3 A x 53.0 ohm = 0.438 V

Hence total voltage required = 0.911 V + 0.390 V + 0.438 V = 1.74 V

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