Redox and redox equilibria

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Chemical cells are the basis of electric batteries. Chemical cells are a practical application of redox reactions. With the help of chemical cells it is possible to define electrode potentials and then use them to predict the direction and extent of change of redox reactions. So this introduces another way of answering the questions: 'How far?' and 'In which direction?'.

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• Key ideas about cells and electrode potentials

Recently Asked Questions

• I am currently investigating the iodine clock reaction involving hydrogen peroxide. I don't understand how to work out if the overall equation is redox, especially in view to oxidation numbers? I also understand that the overall reaction can be split into 'half-equations' but am not sure how to go about this. 210208
• I'm doing my background chemistry for a redox titration to determine the concentration of copper ions using potassium iodide. I know the ionic equation I need is: 2Cu²⁺(aq) + 4I^-(aq) -> 2CuI(s) + I₂(aq) I'm having problems explaining why this reaction occurs though because my two half-equations are: Cu²⁺(aq) + e^- ->Cu⁺(aq) - E^o = 0.15 V I₂(aq) + 2e^- -> 2I^-(aq) - E^o = 0.54 V The E^o values would normally mean I would reverse the Cu half-cell as it has the more negative E^o value. But to get my overall ionic equation I actually need to reverse the iodine half-cell. So I'm confused why the reaction occurs. Vikki
• The oxidation number of vanadium in a complex ion was determined as follows: a 0.013 mol sample of the complex was dissolved in water and the solution made up to 100 cm³. A 10 cm³ portion of this solution required 20.8 cm³ of 0.025 mol dm^-3 KMn0₄ to oxidise all the vanadium to the +5 state. Calculate the original oxidation number of the vanadium. Thank you for teaching me how to solve this problem. 110310

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Tests

• Redox equilibria (A2)

updated: 21 August 2003