Consider the following electrolysis reaction. Cathode: H_2O(l) + e^- reversiblea

Question

Consider the following electrolysis reaction. Cathode: H_2O(l) + e^- reversiblearrow 1/2 H_2 (g, 1.0 bar) + OH^- (aq, 0.01 M) Anode: Br^- (aq, 0.10 M) reversiblearrow 1/2 Br_2(l) + e^- Calculate the voltage needed to drive the net reaction if current is negligible. Suppose that the cell has a resistance of 2.0 ohm and a current of 100 mA. How much voltage is needed to overcome the cell resistance? This is the ohmic potential. Suppose that the anode reaction has an overpotential of 0.20 V and that the cathode overpotential is 0.40 V. What voltage is needed to overcome these effects combined with those of parts (a) and (b)? Suppose that concentration polarization occurs, [OH^-], at the cathode surface increases to 1.0 M and [Br^-], at the anode surface decreases to 0.010 M. What voltage is needed to overcome these effects combined with those of (b) and (c)?

Explanation / Answer

b) Given that Resistance R = 2.0

current I = 100 mA = 100 x 10-3 A = 0.1 A

voltage V = ?

From Ohm's law,

V= IR

= 0.1 A x 2.0

= 0.2 V

Therefore,

voltage required to overcome cell resistance = 0.2 V

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