# Tag Info

5

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!

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 ...

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 ...

3

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. $$\Lambda[\pu{Sm^2mol^-1}]=\frac{M[\pu{g/mol}]}{c[\pu{ppm as g/m^3}]}\cdot ... 2 Oxidation half reaction (happens at the anode)$$\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}$$The electrons are supplied via the wire, and the silver cations leave the ... 2 The voltage you measure between the terminals of a voltaic cell will depend on two factors: The intrinsic maximum voltage (V_\mathrm{max} = E_\mathrm{cell}) that the cell could produce, depending on the E^o_{red} of each half cell, the ion concentrations and the temperature. This is calculated from the Nernst equation:$$E_\mathrm{cell} = E^⦵_\mathrm{...

1

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 ...

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 ...

1

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 ...

1

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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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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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 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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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 ...

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