New answers tagged redox
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If is is dry - nothing. However even in a typical wall temperature changes could possible cause condensation which would preferentially corrode the steel ( "iron" is rare today). I had a house once with emt ( galvanized steel thin wall tubing for electric wiring). Because of an unusual heating situation, moisture condensed in the emt in the walls. ...
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As you state, iron will corrode more rapidly in contact with a more "noble" metal, e.g. copper. That is why plastic pipe hangers ("clamps") may be used. Also, iron pipe hangers are sold with rubber inserts, both to reduce vibration conducted to the wall and to reduce this galvanic corrosion.
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As it is pointed out in other answers and in literature, it is indeed thermodynamically unstable and its reaction synthesis has unfavorable pathways. But, is it "non-existent"? Not quite. There was a possibility of its existence indicated by the fact that hydrated ferric oxide dissolves in hydriodic acid, yielding a brown solution. Its ...
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Iron(III) iodide as a binary salt is highly unstable/transitory, but stable complexes are known with appropriate ligands.
Pohl et al. [1] first synthesized such a complex, $\ce{FeI3(SC(N(CH3)2)2)}$, actually oxidizing iron(II) iodide with elemental iodine in the presence of a carefully controlled amount of $\ce{((CH3)2N)2CS}$. The iron(III) iodide, despite ...
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Thermodynamics courses usually start by calculating amounts of heat and amounts of work enterring a container. And the work is $\pu{p\Delta V}$. No mention of concentration ! Just the pressure. Afterwards, enthalpy is introduced, then gas chemistry is developed, always using pressures. Equilibrium constants are then introduced, always with gases and ...
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I've only basic training in chemistry, since my competences are in Electrical Engineering, but I'd like to tackle your question from another angle.
I'm assuming your question is sort of an X-Y problem and you are not really interested in increasing the efficiency of an actual lemon-based cell.
From an engineering POV, what you want from a power source is, ...
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Yes, you can have same the electrolyte and a pair of two different metals, but the key point is that if you wish to increase voltage difference, you need to connect them in series and use separate containers for each pair, and of course each pair must be connected. Your postulate in the comment is correct. If we use a large bucket, only the pair connected to ...
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Yes, you could have multiple electrodes in the same electrolyte, but to some extent, that would short-circuit the battery. For example, if you stack copper and silver coins with blotting paper (bp) between them, in the order:
Cu bp Ag Cu bp Ag ... Cu bp Ag and immerse the whole in an electrolyte, rather than just wetting each piece of blotting paper, some of ...
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I would solve this problem by algebra, without thinking about oxidation numbers. It is quicker. Let the coefficients be $a, b, c, d, e, f$ in front of the reactants and products. It yields $$\ce{a KSCN + b H2O + c I2 -> d KHSO4 + e HI + f ICN}$$ The number of each atom must be the same on both sides of the equation arrow. Counting atoms by order of ...
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Use molecular species in your half-reactions.
In your case, identify thiocyanate as the reducing agent and, since the cyano function combines with iodine, include enough iodine in that half-reaction to form the iodine cyanide/cyanogen iodide product. This gives
$\ce{SCN^- + (1/2)I2 -> S^{(+6)} + ICN +7 e^-}$
where the +6 oxidation state on sulfur in the ...
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The slow reaction dissolving silver must be : $$\ce{2Ag + H2O2 + 2CH3COOH -> 2CH3COOAg + 2 H2O}$$ The mixture has to be hot because the rection is extremely slow. Its rate is negligible at room temperature. Even at $60$°C, about $10$ minutes are required to dissolve nanoparticles of silver on the milligram scale. Silver acetate $\ce{CH3COOAg}$ is ...
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Please tell your teachers that the concept of chemical equivalents is obsolete. It was taught 50 years ago. Hope your teachers can communicate to the public exam authorities!
$$\ce{CrO4^2- -> Cr}$$
If it were simple molarity, you would take the experimental weight of dichromate (Salt) and convert that into moles by dividing the mass by formula weight of ...
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