6
Unfortunately, the OP's starting reaction expression is simply wrong: the left hand side has only a neutral compound while the right hand side has anions among the product species. From the wikipedia article on fluorine nitrate, the unbalanced reaction of fluorine nitrate and water is
$$\ce{a FNO3 + b H2O -> c O2 + d OF2 + e HF + f HNO3 \tag 1}$$
For ...
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 ...
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 ...
2
If you start with $\ce{CO2}$, you could have one reaction that yields $\ce{CO}$ and $\ce{O2}$, and another where $\ce{O2}$ dissociates:
$$\ce{2CO2(g) -> 2CO(g) + O2(g)}\tag{1}$$
$$\ce{O2(g) -> 2O(g)}\tag{2}$$
Then, you can combined these two in arbitrary ways, or like this with a variable $x$:
$$\ce{2CO2(g) -> 2CO(g) + (1-x) O2(g) + 2x O(g)}\tag{1+2 ...
2
As noted elsewhere, the reaction as given is not balanced. Assuming that other products such as $\ce{OF2}$ are ignored, apply the following rules for basic solution:
Use hydroxide ion to allow for balanced charges.
Use water to allow balancing hydrogen and oxygen atoms.
Here, hydroxide ions should he added on the left by Rule 1 and then water is needed ...
1
I have more succinctly arrived at Eq(9) as proposed by Ed V above, by noting observed reactions as reported in the literature.
For the record, I do not recommend working with FNO3, a shock-sensitive explosive. My expected action of FNO3, with say dry KOH, is better described simply by the term energetic. Working with dilute aqueous alkaline solutions is ...
1
I agreed with EdV about OP giving an errotic equation. Since $\ce{NO3-}$ remained unchanged in RHS of the equation, we need a compound to reduce $\ce{F+}$ to $\ce{F-}$ in this redox reaction. Since $\ce{O2}$ is being a product, we can choose one of the following half reactions to do the trick based on the medium (acidic or basic):
$$\ce{O2 + 4H+ + 4e- <=&...
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
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
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 ...
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
$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$
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 ...
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