New answers tagged

1

The Cannizzaro reaction is the disproportionation of an aldehyde, which do not have an $\alpha$-hydrogen, to an equimolar mixture of corresponding primary alcohol and carboxylic acid salt. For example, original Cannizzaro's experiment: $$\ce{2C6H5-CHO ->[aq NaOH] C6H5-CH2OH + C6H5-COO^-Na+}$$ Since there is no $\alpha$-hydrogen, the mixture does not ...


1

Some electrolysis experimets are best first performed and then an attempt at explanation. For example, the products of the electrolysis of aqueous nickel nitrate with graphite electrodes are reportedly per a source (with video) as: Electrolysis of a nickel nitrate solution produces oxygen at the anode, and hydrogen and nickel at the cathode. Looking at ...


1

$\pu{0.7542 eV}$ as calculated by Mills (Ref.1) and measured experimentally by Lykke, et al. (Ref.2). This is much less than the well-known $\pu{13.6 eV}$ ionization energy of Hydrogen and follows the general rule that ionization energies increase as you remove more electrons. In this case, the positive charge of the nucleus is entirely screened by the first ...


1

One aspect to consider is that the boiling points of alkali metals go down as they get heavier. From The Encyclopedia Britannica: Lithium: bp = 1342°C Sodium: 883°C Potassium: 759°C Rubidium: 688°C Cesium: 671°C Francium: 677°C (probably this one is merely predicted) Thus the heavier alkali metals could be volatile enough to effectively escape as ...


-2

This seems to be a thought experiment, and many factors have been considered, but not enough to correspond to reality. One item to consider is that copper is electrolytically purified by psssing current between a thin cathode of pure copper and a thick cathode of impure copper. An electrical power supply drives the copper (in solution) from the anode (+) to ...


2

When you dip magnesium ribbon into a copper sulfate solution, theoretically (and realistically) you do get a simple replacement reaction. Then reality sets in: you have a metallic anode (magnesium) with little cathodes (copper) all over it. At that point, the magnesium just overreacts. Well, it reacts faster. The size of the bubbles will cause some ...


3

Yes, the standard reduction potential does depend on temperature. The definition of the standard reduction potential is stated in Ref.1 as: A standard electrode potential $E^\circ$ is defines as the potential (in Volts, $\pu{V}$) of a half-reaction relative to a reference electrode at a specific temperature, all chemicals being at their standard states at ...


0

I've also searched multiple places for the answer and came across several similar questions. What causes electrons to move from zinc to copper? Why do Electrons leave the Zinc in a Galvanic Cell However, none of the answers inside seems to be on point or understand the intention of the people who asked the question. I've thought about the problem and have an ...


1

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $...


1

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$ Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $\ce{Na3PO4}$ and at the right hand side 3 moles of $\ce{NaCl}.$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $\ce{Cl-}$ ion. Hence $n\text{-factor} = 3.$ In reactions like this where there's no change in ...


2

Bromous acid ($\ce{HBrO2}$) is an inorganic compound, which is an unstable compound. However, the salts of its conjugate base, bromites (e.g., $\ce{NaBrO2.3H2O}$) have been isolated. Bromous acid is considered to be an oxidizer. However, I can't find its reduction potential, probably because of its unstable nature (Ref.1). Based on the studies of Faria, et ...


2

If you need a good answer, you should not put an information embargo on the context of your question. This is to be taken rather as an expanded comment than a full answer: $\ce{HBrO2}$ oxidizes $\ce{Br-}$ to $\ce{Br2}$. $$\ce{HBrO2 + 3 Br- + 3 H+ -> 2 Br2 + 2 H2O}$$ To tell if it may compete with $\ce{Br-}$ to react with the "chemical", we ...


1

If there is no bridge in a voltaic cell, this will be no cell any more. The arrangement will just be a metallic connection between two different containers. It will not deliver any continuous current. And replacing the salt bridge by a wire will simply add a second metallic connection between the solutions. There will be two parallel but different metallic ...


0

To quote an educational source: Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water. which means, in essence, that aqueous NaCl is a good ...


Top 50 recent answers are included