Please help with this question if you can! :) The first answer .009 is wrong...

Question

Please help with this question if you can! :) The first answer .009 is wrong...

Map Sapling Learning Consider the following electrolysis reactions Cathode Cur (aq, 0.234 M)+ 2e Anode Holl o (g, 1 bar) 2H (aq, 0.166 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 .009 The cell has a resistance of 67.0Q and a current of 7.52 mA when the cathode overpotential is 0.310 V and the anode overpotential is 0.058 V. Calculate the voltage needed to overcome these effects and drive the net reaction. Number

Explanation / Answer

Calculate

E°cell = Ered - Eox

E°cell = -1.23 - 0.34 = -0.89 V , negative for electrolysis

so

overal reaction

Cu2+ + H2O --> Cu(s) + 1/2O2 + 2H+

so

[Cu+2] = 0.234

[H+] = 0.166

PO2 = 1 bar

so...

Ecell = E°cell - 0.592/n*log(Q)

Q = [Cu+2]/([H]^2[PO2)

substitute

Ecell = -0.89 - 0.0592/(2) * log(0.234 / (0.166^2)*1)

Ecell = -0.91749V

So..

the current required must be + 0.91749 V

Q2

cell overpotential = Cathode Overpotential + Anode overpotential = 0.310 + 0.058 = 0.368 V

The V for

7.52 mA and R = 67

V = IR

IR = (7.52*10^-3)(67) = 0.50384 V

so

Vtotal = 0.91749 + 0.50384+0.368 = 1.78933 V

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