(a) Sketch a cell with inert electrodes suitable for the elec-trolysis of an aqueous solution of sulfuric acid. Label the anode and cathode, and indicate the direction of electron and ion flow. Identify the positive and negative electrodes. (b) Write balanced equations for the anode, cathode, and overall cell reactions.

Video transcript

Hello everyone in this video. We want to go ahead and draw the cell setup of the electrolysis of aqueous hydrochloric acid. We have to go ahead and identify the anodes and cathodes and indicate the direction of flow of electrons and read out the balanced overall self reaction. So let's go ahead and write or squeeze this in. So for my cathode side, which I'll go ahead right out in red we have two moles of my H plus in its agree estate that reacts with two electrons and this yields are H two in its gaseous state. The standard reduction potential here is equal to zero volts Now. Over on my own. Outside, which I'll do in blue, we actually have two half reactions going on first, we'll have these two moles of our cl minus and it's a great state. This yield to give us cl two in its gaseous state as well as two electrons. The standard reduction potentials is equal to 1.358 volts. My second half reaction that occurs in this is our two moles of our H 20 in its liquid state. And this yields 02 and its cash estate as well as four H plus And our four electrons. And the standard reduction potential here is equal to 1.229V. So This half reaction here is not preferred because of the over potential from the formation of our 02 guests and knowing this then we can write out the overall reaction which will do over to the left in purple. The overall reaction that we get here is that we have two moles of H plus As well as two moles of RCL in this book. A hen yield our H two and ask ASHA state as well, Cl two and ask ASHA state. The sl value that we get here is equal to negative 1.358V is what we get by in both of our center reduction potentials from the selected half reaction at the cathode and an ode. So this means that we have an electronic cell because we have this negative value and this type of cell, the unknown is the positive terminal. While the cathode is the negative terminal, electrons will flow from the nose to the battery to the cathode. So I have all my answer tresses ranging from A, B, C and D. The one that best matches not only our calculations and our finalization of the data, it's going to be answer choice A. So as a choice A here is going to be my final answer for this problem