Porous pellets of TiO₂ can be reduced to titanium metal at the cathode of an electrochemical cell containing molten CaCl₂ as the electrolyte. When the TiO₂ is reduced, the O^2-ions dis-solve in the CaCl₂ and are subsequently oxidized to O₂ gas at the anode. This approach may be the basis for a less expensive process than the one currently used for producing titanium.

(a) Label the anode and cathode, and indicate the signs of the electrodes.

Video transcript

Hello. In this primary told small scale production of strong metal can be performed using strontium chloride placed an electrochemical cell with molten potassium chloride as the electrolyte. Upon reduction of strontium chloride chloride ions are subsequently oxidized chlorine gas set up this picture below, we are asked to identify the reactions of the cathode and the anodes. Given that the problem states that this is for small scale production of strong metal, we might just go ahead and assume then that the reaction that's taking place at the cathode is the reduction of strontium ions. So that's considered though other reactions that might take place at the cathode. So we have strong minds being reduced in form strong and metal Standard reduction potential is equal to negative 2.899V. And then we could also have potassium mines being reduced to form potassium metal. The standard reduction potential for this half reaction is negative 2.931V. And so when we look at these two standard reduction potentials, we see that the one for strontium being reduced from strontium metal has the less negative standard reduction potential. That means then the strontium has a greater potential to be reduced as compared to potassium. So the reduction of strontium will be the reaction that takes place at the cathode. Now if we consider the anodes. So at the anodes is where oxidation takes place. So we will have a two chloride irons than being Oxidized to form foreign gas standard oxidation potential associated with this reaction is negative 1.358V. And so this will be then the half reaction that takes place at the anodes. And this then corresponds to answer. Be Thanks for watching. Hope. This help.