New answers tagged redox
-1
If you replace sulfuric acid by a sulfate, the half-equations will become : $$\ce{Pb + SO_4^{2-} -> PbSO_4 + 2 e^-}$$ $$\ce{PbO2 + SO4^{2-} + 2 H2O + 2e- -> PbSO4 + 4 OH-}$$
As a consequence, the solution will loose sulfate ions which will be replaced by $\ce{OH-}$, which is unfavorable, whatever the origin of the ion sulfate, as will be shown now.
...
-1
Disagree as your selected salts are not sources of the $\ce{HSO4- ion}$, as is cited in both half-cell reactions occurring in the Lead-Acid Battery. For example, per Wikipedia, to quote:
Negative plate reaction
$$\ce{Pb(s) + HSO4-(aq) → PbSO4(s) + H+(aq) + 2e− }$$
And further:
Positive plate reaction
$$\ce{PbO2(s) + HSO4-(aq) + 3H+(aq) + 2e− → PbSO4(s) + ...
2
Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving ...
1
By definition : $\ce{\Delta $G = - zEF$}$, where $z$ is the number of electrons in the half-reaction, and $F$ is the Faraday (about $96'500$ Cb)
In the first half-reaction , $z = 1$, and $E = +~ 0.98$ V. So $\Delta G_1 = - ~0.98~ F$ (in Joules)
In the second half-reaction, $z = 2$, and $E = +~1.59$ V. So $\Delta G_2$ = - $2·1.59 F$ = - $3.18 F$ (in Joules)
...
2
Dichromate (orange) is present in acidic medium while chromate (yellow) prevails in basic medium. Here are stepwise solutions in both media. Acidic Medium:
Recognizing that chromium is the oxidant, balance the chromium atoms.
$$\ce{Cr2O7^2- -> 2Cr^3+}$$
Because the reaction is conducted in aqueous acid, only water and protons are available to balance the ...
0
In this reaction, the ion $\ce{Ce^{4+}}$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $\ce{Ce^{3+}}$. This electron may come from the central function $\ce{-CHOH-}$ of any secondary alcohol $\ce{R-CHOH-R'}$. The bond between $\ce{O}$ and $\ce{H}$ is able to loose $1$ electron and $1$ $\ce{H+}$ ion. Same thing ...
2
Let's get the simple question out of the way first:
How do you balance a redox reaction lacking a molecular reactant on the product side? I am referring to the atom or molecule that changes oxidation state when I say molecular reactant.
You can not balance it then. Also, you wouldn't be given such a reaction and is always a typo when they've said that it ...
4
In reality, benzene and acetaldehyde are the minor product of the reaction of benzenediazonium chloride with ethanol. The major product is ethoxybemzene (phenetole). Peter Griess, who had discovered diazonium salts in 1858, had reported benzenediazonium salt (nitrate or sulfate) with ethanol undergoes aforementioned redox reaction in 1864 (Ref.1). However, ...
1
Metallic aluminum cannot be obtained from a solution of its salts. It is because pure aluminum reacts with water, a little bit like sodium. Of course a piece or a sheet of aluminum does not usually react with water, because the metal is covered by a thin, colorless and waterproof layer of aluminum oxide, which prevents the contact between aluminum and air. ...
3
I want OP to understand that iodine ($\ce{I2}$) is not the one giving the starch solution deep blue color. It is rather triiodide ion ($\ce{I3-}$), which would form in presence of excess iodide ion ($\ce{I-}$):
$$ \ce{I2 + I- <=> I3-} \tag1$$
This is an important aspect in iodometry titrations, because $\ce{I3-}$ is very water soluble compared to ...
6
Starch is a long ribbon or a long filament made of a great number of glucose units attached to one another like the wagons in a train. But this filament is wound in a helicoidal way. And the inner part of this helix is a sort of long hole. And the dimension of this hole is just big enough to allow the iodine molecule $\ce{I2}$ to enter and stay there, ...
1
Looking at the standard electrode potentials answers your question (Note that conc. $\ce{HNO3}$ oxidizes $\ce{Pb2O3}$ to $\ce{PbO2}$) :
$$\begin{align}
\ce{PbO2 + 4H+ + 2e- &-> Pb^2+ + 2H2O } & E^0 = \pu{1.46 V} \\
\ce{S2O8^2- + 2e- &-> 2SO4- } & E^0 = \pu{2.01 V} \\
\ce{SO4^2- + 4H+ + 2e- &-> SO2 + 2H2O } & E^0 = \pu{0.17 V}
...
10
What you have written is right, but it is just the first step.
The release and solvation of an electron is fast. Reaction of an electron with ammonia to release hydrogen $$\ce{2 e-(solv) + 2 NH3 -> H2 + 2 NH2-(solv)}$$ is slow. Deep blue (diluted) or bronze (concentrated) "sodium electride" solution in liquid ammonia slowly converts itself to ...
1
@Sir Arthur7 did a fine job of discrediting D-ketose (d) as a viable candidate for structure A. Belatedly, I will expand upon my comment above at the request of @Jan. [In my comment above, "glutaric" should read "glucaric".]
Assuming that $\ce{HNO3}$ is capable of tautomerizing the D-ketose via the enediol and accomplishing terminal ...
1
Just to add some more information for you with some clarification. I gather that you are asked to calculate the equivalent weight of nitric acid here.
$$\ce{Cu + HNO3(aq, dil) -> Cu(NO3)2 + NO(g) + H2O(l)}$$
So first thing to keep in mind that the equivalent weight is determined with respect to a single role of the molecule of interest. An exception is a ...
0
As you noted, IUPAC does discourage using the equivalent weight concept because the value is fuzzy. Consider the following reactions:
Dissociation of Nitric acid
$$\ce{HNO3 ->[aq] H+ + NO3-}$$
Reduction of Nitrate acid to Nitrogen dioxide
$$\ce{NO3^- + 2H+ + e- ->[aq] NO2(g) + H2O}$$
Reduction of Nitrate acid to Nitrogen Monoxide
$$\ce{NO3^- + 4H+ + ...
4
Please forget about equivalents and normal solutions ! It is an old theory, that was abandoned in the middle of the $20$th century. It has been replaced by moles, molarity, and similar parameters. A solution containing $n$ moles of a solute in one liter water has a concentration expressed in mole per liter, that can be printed on the label of the flask. The ...
0
For your future assignment it is always worth work with two half-reactions: oxidation and reduction: Given is:
$$\ce{NiO2(s) + 4 H+(aq) + 2 Ag(s) -> Ni^2+(aq) + 2 H2O(l) + 2 Ag^+(aq)}\quad
E^\circ = \pu{2.48 V}$$
Oxidation half-reaction:
$$\ce{Ag (s) <=> Ag+ + e-} \tag1$$
Reduction half-reaction:
$$\ce{NiO2 (s) + 4 H+ + 2e- <=> Ni^2+ + 2H2O (l)...
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