New answers tagged

-1

A proton combining with an electron doesnt necessarly release energy.All the posts forget that there is the weak and strong interaction.For example let's take hydrogen atom.If the electron from the 1s orbital is combined with the proton(core) it will leave a neutrally charged neutron.But neutron is unstable when it isnt in the core of elements , so it will ...


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If the process of uniting protons and electrons is assumed to occur isothermically in a vessel of constant volume, then the heat released will be equal to the change in internal energy of the system: $$q_v=(\Delta U)_v$$ Now if the mixture were to behave as an ideal gas, the change in internal energy would be zero and therefore so would the heat. But we ...


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One way to think of it is the electrical potential energy of the separated charges is released (as heat energy, eventually) when the charges unite. First the energy will be released as photons, and these photons will be absorbed by nearby molecules, causing them to increase in kinetic energy, which on the macro scale, is heat.


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You can see my detailed response on the sign of electrode potentials here: Still taught to reverse oxidation half cells in electrochemistry? To summarize, electrochemists all over the world have decided to quote all electrode potentials as reduction potentials. This tug of war of signs between the so-called European convention and American convention of ...


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Single electrode reactions are possible if you do electrostatic electrochemistry. Allan J Bard and his students published a paper in Nature. They electrostatically charged a Teflon and dipped in copper (II) solution. The result was the reduction Cu(II) + 2e = Cu(0). It is a detailed paper, have a look at


1

Both answers are very good, but Mathew Mahindaratne's answer uses a galvanic cell figure that needs some work. I respectfully request that his galvanic cell figure be replaced and I hereby offer up my handmade figure (see below), either as a temporary filler or for whatever period Mathew may decide. It is entirely up to him. I do not claim my handmade figure ...


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A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. An electrochemical cell consists of two electrodes (anode and cathode), two electlolyte solutions, and a salt bridge: The anode is an electrode where oxidation occurs and the cathode is an electrode where reduction occurs. The oxidation and reduction ...


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Oxidation half reaction (happens at the anode) $$\ce{Mg -> Mg^2+ + 2e-}$$ The electrons get removed via the wire, and the magnesium cations enter the aqueous solution, giving it a net positive charge. Reduction half reaction (happens at the cathode) $$\ce{Ag+ + e- -> Ag}$$ The electrons are supplied via the wire, and the silver cations leave the ...


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Corrosion is a more or less involuntary/undesired reaction. If you apply a current, the corrosion is defined by the current you apply, so you are controlling the system, not just observing it. In general, electrochemical forcing puts your system into a highly unnatural regime, but can certainly provide interesting information. Corrosion can be affected (in ...


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In the galvanic cell you are talking about, the copper is the inactive electrode. The activity, the electron production, comes from the more active metal. When a neutral metal atom falls into solution, it will leave is electrons behind: $$\ce{M -> M^2+ + 2 e-}$$ But those electrons are just a few of the many that are left behind by all the $\ce{M}$ ...


3

The equation you quote is the empirical relationship discovered by Kohlrausch (1907) for the conductivity of strong electrolytes, and is valid for concentrations up to $\sqrt{c} \approx 0.3$. $\Lambda_m$ is the molar conductivity usually given with units $\mathrm{\Omega^{-1}\,cm^2\,mol^{-1}}$ ($\Omega$ = ohm). The $\Lambda_m^0$ is the molar conductivity at ...


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This is pure scam! All he is doing is using an electrode which is corroding during electrolysis. The flocculated material you see in tap water is just a metallic hydroxide. If tap water had this much of metals, nobody will survive. Such a fraudulent salesman should be shooed away!


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The galvanic cells from the same material are possible. They are called concentration cells and they use two electrolytes of the different concentration of the same electrolyte. But their voltage is low. In the case of $\ce{CuSO4}$ $$U = 0.0295 \log\frac{c_1}{c_2}\ \mathrm V$$ Note that the electrode naming is based on the direction of the electron flow, ...


1

It's tempting to use phase separation with an aqueous phase, but you're clearly having issues. One option is to evaporate the volatile solvents off, add toluene, cyclohexane or heptane and try water or another additive to boost the ionic strength. The approach using NaOH suggests this may not be successful. A less conventional approach may be to partition ...


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What is the "correct" answer You are asking why the answer key has water and not saltwater as an answer. Decomposition is defined as: A reaction where a single compound breaks down into simpler compounds. For salt water, the salt reacts at one electrode, and the water at the other, and you are making NaOH. For pure water, you are splitting water into ...


4

Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge. $$\ce{Pb <=> Pb^2+ + 2 e-}$$ At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$. $$\ce{Pb^4+ + 2 e+ <=> Pb^2+}$$ For a classical lead–acid battery, the overall ...


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ppm is ambiguous quantity, it can be v/v, w/v,w/w,n/n. By other words, all 1 ppm very probably is not the same molar concentration, unless it is n/n ppm, relating itself to the molar amount of the solution. $\Lambda = \kappa /c$ is correct, but $c$ must be the molar concentration. $$\Lambda[\pu{Sm^2mol^-1}]=\frac{M[\pu{g/mol}]}{c[\pu{ppm as g/m^3}]}\cdot ...


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This differs from the lemon galvanic cell (aka, 'battery') in several important ways: 1) Mg is used as the anode, rather than Zn, 2) there is no salt bridge, unlike in the the lemon cell, where the lemon serves as a crude 'salt bridge', and 3) the electrotyle is 2 M HCl rather than citric acid (and other stuff) in lemon juice. An open circuit voltage will be ...


1

It does, as an intermediate product. The peroxide formed is on the surface of the electrode which changes it's chemical behavior. Electrolysis is a four-electron process. In the first step, an electron is removed from a water molecule, producing a neutral OH adsorbed to the surface of the electrode. The second step removes another electron from the adsorbed ...


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The electrolyte in that clip is a solution of $\ce{NaCl}$ and acetic acid (vinegar). The zinc anode is oxidized, resulting in $\ce{Zn^2+}$ ions going into the electrolyte. The electrons went through the red LED, lighting it, and then on to the copper cathode. From there, they reduced some $\ce{H+}$ to hydrogen gas. So two $\ce{H+}$ ions were reduced to one $\...


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It is the same principle as why there is not going any current from a voltage source to a powered device, if there is currupted galvanic connection anywhere. The initial current very quickly balances potentials of power outlets with potentials on the wires and no current is flowing after charging the wire parasitic capacitance. Let consider the classical ...


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My question is that $E^\circ$ is just the potential at some given standard condition, i.e. 1 bar and 25 °C.... No, $E^\circ$ is the potential at the given standard condition 25 °C, 1 ATM and activities of involved compounds equal to 1. This may be the source of your confusion. Standard potentials are for the unit activities. In my example, once for silver ...


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The voltage you measure between the terminals of a voltaic cell will depend on two factors: The intrinsic maximum voltage $(V_\mathrm{max} = E_\mathrm{cell})$ that the cell could produce, depending on the $E^o_{red}$ of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation: $$E_\mathrm{cell} = E^⦵_\mathrm{...


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