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Postby JChen_2I » Fri Feb 28, 2020 3:18 pm

A 1.0 m NiSO4(aq) solution was electrolyzed by using inert electrodes. Write (a) the cathode reaction; (b) the anode reaction. (c) With no overpotential at the electrodes, what is the minimum potential that must be supplied to the cell for the onset of electrolysis?

How do you find the minimum potential? The solution manual uses 2H2O->O2 +4H+ +4e- (E=+1.23) as the anode reaction. Why would you not use 2H2O +2e- -> H2 + 2OH- (E=-0.42)?

Alex Chen 2L
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Re: 6O1

Postby Alex Chen 2L » Fri Feb 28, 2020 4:25 pm

In electrolysis, you are trying to produce solid, neutral Ni from its ion form as Ni2+ in NiSO4(aq). To do this, you will need to supply electrons from water to reduce the Nickel ions (water is oxidized and electrons are in the products of the half-reaction), which becomes the anode reaction. Therefore Ecell is the standard E of the cathode (nickel) minus the anode (water), -0.23 - (1.23) = -1.46V. Since the Ecell is negative as a result and unfavorable in that direction, the minimum potential required to drive the reaction is the amount of volts added to make Ecell zero (or simply negative Ecell).

Radha Patel 4I
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Re: 6O1

Postby Radha Patel 4I » Sat Feb 29, 2020 12:12 pm

In Table 6M.1 the reaction is written as the reduction reaction O2 +4H+ +4e- --> 2H2O with E ^o= +1.23V. In the answers they wrote it as the oxidation reaction by flipping the reaction to 2H2O->O2 +4H+ +4e- (E=+1.23). When we flip the reaction, the value of E^o becomes negative. When we pick an anode reaction, we choose the oxidation reaction that is more negative. This is why they chose the reaction with E=+1.23V instead of 2H2O +2e- -> H2 + 2OH- (E=-0.42) because -1.23V is more negative than -0.42V. As for the maximum potential the E^o cell is -1.46V so the minimum potential must be +1.46 since 1.46-1.46 equals 0. E(supplied) must be above 1.46V to drive the reaction. This is how I interpreted this question, but I'm not 100% positive if I am correct.

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