Determine Electrolysis Products via E Value

Let's predict the electrolysis products of dilute KOH using platinum electrodes. The species to consider are: Cathode: K+ and water Anode: OH- and water 1. Reduction at Cathode Since K+ and water can be reduced at the cathode, we have to use their electrode potentials or E value to compare their ease of reduction. Species reduced are found on the left hand side of the half equations in the Data Booklet, so we focus on finding K+ and H2O on the left hand side of half equation. Water has a more positive E value, more likely reduced hence will be reduced at the cathode. Product at cathode is H2 gas. 2. Oxidation at Anode OH- and water can be oxidised at the anode and we can use E value to compare their ease of oxidation. Species oxidised are on the right hand side of half equation, so we focus on finding OH- and H2O on the right hand side of half equation. OH- has a more negative E value, more likely oxidised hence will be oxidised at the anode. Product at anode is O2 gas. Topic: Electrochemistry, Physical Chemistry, A Level Chemistry, Singapore Please SUBSCRIBE to my channel and SHARE this video with your friends if you find it helpful! View the full video with description at my website https://chemistryguru.com.sg/determine-electrolysis-product-via-e-value This video is extracted from our Chemistry Guru on-demand video lessons. Learn A Level Chemistry online anytime, anywhere and at your own pace! Find out more at https://chemistryguru.com.sg/on-demand-video-lesson