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If I want to increase the voltage, which electrolyte do I change? Hope you are familiar with the term electrolyte which is the solution where ions flow in the electrochemical cell The electrolyte solutions used here are $\ce{ZnSO4(aq)}$ solution and $\ce{CuSO4(aq)}$ solution The anode or cathode? Yes you can chose both or one of them Now lets just ...


2

May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes. image and further information here


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As suggested in the comments, the key is Nernst equation. The fundation is the Gibbs free energy given by: $\Delta G = \Delta G^0 + RT \ln Q_r$ where $\Delta G^0$ is the standard free energy for the process, while $Q_r$ is the reaction ratio. Remember that the reaction ratio is the analogous quantity of the equilibrium constant, but with the non equilibrium ...


3

I want to elaborate the answer of Ashish Ahuja because he is also perplexed by the n-factors. First of all for general readers n-factors are of historical interest only. Due to my interest in historical analytical methods, the following points would clarify this situation: A) Equivalent weights do not require any knowledge of molarity, moles, electrons ...


3

We can split the reaction $$\ce{H2O2 -> H2O + O2}$$ into the respective reduction and oxidation half-reactions. $$\ce{H2O2 -> O2 + 2H+ + 2e-}$$ $$\ce{H2O2 + 2H+ + 2e- -> 2H2O}$$ Since the n-factor of $\ce{H2O2}$ for both these half-reactions is 2, the n-factor is: $$\frac{1}{n_f} = \frac{1}{2} + \frac{1}{2} = 1$$ $$n_f = 1$$ Based on the comments ...


1

First put your redox potentials on a horizontal line by order of increasing standard redox potentials. You will see that it starts on the left-hand side by $\ce{Zn^{2+}/Zn}$ at $\pu{-0.76 V}$. Then, going to the right, you find $\ce{V^{III}/V^{II}}$ at $\pu{-0.26 V}$. Now you find $\ce{Pb^{2+}/Pb}$, and so on. Now try to learn how to use this scale. The ...


2

The solubility product of $\ce{PbSO4}$ is about $10^{-8}$. Now suppose $[\ce{Pb^{2+}}$] falls down from $\pu{1 M}$ to an arbitrary low value like $\pu{10^{-8} M},$ due to addition of $\ce{SO4^{2-}}$ ions. In this case, Nernst's law can be applied, and the potential of the lead electrode falls from $E^\circ_\ce{Pb} = \pu{-0.13 V}$ down to $$E_\ce{Pb} = \pu{-0....


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There is oxidized oxygen in water $$\ce{2 H2O -> 4 H+(aq) + O2(g) + 4 e-}.$$ $\ce{HSO4-}$ in sulphuric acid solution can get oxidized as well at high current density and therefore high enough anodic potential: $$\ce{HSO4-(aq) -> S2O8^2-(aq) + 2 H+(aq) + 2 e- }$$ This is the industrial way to produce potassium peroxodisulphate.


1

Let's see if we can tease out the main points here. The images below are taken from Lance and Cole's Analytical Chemistry POGIL workbook. Microscopic processes at a metal surface exposed to an aqueous solution containing the corresponding metal ion(1) will result in a slight negative charge on the metal surface due to oxidation of the metal and a slight ...


0

Theoretically, we could do any reaction. Practically, some are only possible in one direction; others are under special conditions. For the case of copper: the standard potential of the pair $ \ce {Cu^{2 +} / Cu} $ being positive, this means that naturally the reaction is done in the direction $ \ce {Cu^{2+} -> Cu} $. But we can still force the reaction ...


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the standard potentials are given in the literature generally in an acidic medium at pH = 0. In this case, we must therefore write the equations in an acidic medium with $ H ^ + $ and not with $ HO ^ - $


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This diagram, called E-pH diagram or Pourbaix diagram, makes it possible to know the domains of existence or of predominance of redox species as a function of pH. The abscissa is the pH, the ordinate the reduction potential (with respect to the standard hydrogen electrode): thus, it is not a question of the size epsilon but of the potential $E$ which has ...


4

This is a Pourbaix Diagram (I assume this is where you got this figure). While it's not a common notation as far as I'm aware, $\epsilon_0$ is being used to represent the reduction potential in volts (V). As mentioned at the top of the Wikipedia page, this is basically a phase diagram, that tells you the ranges of voltage/pH for which a particular species of ...


2

To be said at first: There is no real displacement. There is an equilibrium reaction of a weak acid/base and their salts, that form mutually a conjugate pair acid/base ( like acetic acid/acetate or ammonium/ammonia ). All is then matter of chemical equilibrium. The stronger acid is, the more is it's dissociation equilibrium in favour of its anion and vice ...


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