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21

The trend in the reducing power of the alkali metals is not a simple linear trend, so it is a little disingenuous if I were to solely talk about $\ce{Li}$ and $\ce{Cs}$, implying that data for the metals in the middle can be interpolated. $$\begin{array}{cc} \hline \ce{M} & E^\circ(\ce{M+}/\ce{M}) \\ \hline \ce{Li} & -3.045 \\ \ce{Na} & -2.714 \...


16

The electronic configuration has nothing to do with it. The reduction potentials of $\ce{Ni^3+}/\ce{Ni^2+}$, $\ce{Cu^3+}/\ce{Cu^2+}$ and $\ce{Zn^3+}/\ce{Zn^2+}$, if they have been/could be measured, would be even greater. The reduction potential for $\ce{M^3+}/\ce{M^2+}$ is most dependent upon the third ionisation energy. If $I_3$ is large then it will be ...


12

Look at the unit volt. 1 volt = 1 joule per coulomb. If the number of moles is doubled, the coulombs are doubled, as is the number of joules. Think of the volt as the driving force behind an individual electron in an electrochemical system.


10

Why Do Carbonates, Oxides, and Pure Metals Precipitate Before Metal Sulfides? They don't. I cannot get to the particular paper you read, but I can provide both a logical and a quantitative argument that the metal sulfides precipitate first. The thermodynamics of this situation do not involve reduction potentials so much as simple solubilities. Metal ...


10

Yes, you can "convert" this way, but you're correct to be skeptical. Let's start with interpreting the cyclic voltammetry curves themselves. (figure from Wikipedia) Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved,...


8

The short answer is thermodynamics. Reduction with $\ce{Cr^2+}$ must be more exergonic than reduction with $\ce{Fe^2+}$, we'll get to some numbers in a bit, but let's deal with the concept. It is tricky to compare the "stability" of two possible products that occur from different pathways. What is more important for the spontaneity of the reaction is the ...


7

Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law. How so? Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new ...


7

For both half-reactions, the actual potentials depend on $\mathrm{pH}$; however, the values given in the question apply to different $\mathrm{pH}$. The given reaction of hydrogen peroxide and the corresponding potential apply to $\mathrm{pH=0}$ or $\left[\ce{H+}\right]=1$: $$\ce{O2 + 2 H+ + 2e- <=> H2O2}\qquad E^\circ=0.695\ \mathrm V$$ Whereas the ...


7

It would really help if you copied the entire half cell reaction. The atoms must balance in a valid half cell reaction! So there are: \begin{align} \ce{NO3− (aq) + 2 H+ + e− &<=> NO2 (g) + H2O} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg2^2+ + 2 e− &<=> 2Hg (l)} &\quad &E_0 = \pu{+0.80 V} \\ \ce{Hg^2+ + 2 e− &<=> Hg (...


6

The cell potential is related to the change in Gibbs free energy by this equation: $$ \Delta G^\circ= -nFE^{\circ}_\text{cell} $$ where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant. The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction ...


6

According to this Wikipedia entry: "[electronegativity] is a chemical property that describes the tendency of an atom to attract electrons (or electron density) towards itself. then according to an electronegativity scale called the Allen electronegativity: "...neon has the highest electronegativity of all elements, followed by fluorine, ...


5

Short answer: Yes. Take a look at the Ellingham Diagram. The line for aluminium oxide is lower than the line for the all the iron oxides. This indicates greater stability of aluminium oxide, or in other words indicates aluminium's greater affinity for oxygen. Thus, based on thermodynamics alone one would expect aluminium to be able to reduce iron oxides. ...


5

For the acidic electrolysis, use the reactions where $\ce{H+}$ occurs. As $\ce{OH-}$ is not available in considerable amount there as a reagent, neither it is created as a product. Generally, for a reaction choice, apply the principle of availability and stability, allowing for a reagent to exist in (relative) abundance. $\ce{OH-}$ or anions of ...


4

Let's consider the following redox couples: $\ce{Cu^{2+} + e^- ->Cu^+\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_1}(\ce{Cu^{2+}/Cu^+}=0.17\, \ce{V})$ $\ce{Cu^{2+} + e^- + I^- ->CuI\,\,\,\,\,\,\,\,\,\,\,\,}{ E^o_2}(\ce{Cu^{2+}/CuI}=?)$ As the chemical species $\ce{Cu^{2+}}$, $\ce{Cu^+}$ and $\ce{CuI}$ are together in the aqueous solution, ...


4

To decide which is the best reducing agent we only not consider that who has less ionisation energy yet it follows 3 steps: Metal (solid) to Metal (gaseous state) sublimation energy Metal from gaseous state to M+ ionisation energy M+ to M+ (aqueous state) hydration energy Lithium having more charge density has more sublimation energy and ionisation ...


4

The answer is hard-soft acid/base theory (or HSAB theory). Essentially, some Lewis acids and bases are hard and some are soft, with a few borderline cases. Hard acids and bases tend to have Small atomic/ionic radius Highly charges (or highly positive/negative oxidation state) High electronegativity for bases and low electronegativity for acids Low ...


4

Galvanizing isn't just a protective coating in the sense that it avoids contact between the iron and the oxidant. Zinc oxidizes more easily than iron, and since both conduct electricity, you can have zinc in contact with the iron anywhere, and it will oxidize first. The term for zinc in this case is "sacrificial anode." Full coating does help additionally ...


4

You used all the information from the question except the line that would have helped you solve it The concentrations of iron-containing species satisfy the relations $[\ce{Fe^2+}] = [\ce{Fe(CN)_6^4-}]$ and $[\ce{Fe^3+}] = [\ce{Fe(CN)_6^3-}]$ From your first cell you find that $$\frac{[\ce{Fe^2+}]}{[\ce{Fe^3+}]} = 103.97$$ which means that in your ...


4

Perhaps your observation is an indication of irreversible reaction conditions. More specifically, a tight deposit on the electrodes hampering the advancement of the electrochemical reactions. This happens sometimes with some less soluble organic dyes used organic electronics, too. Take a look if the electrodes are still shiny, maybe they need to be cleaned ...


4

Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1, If we consider reactions $$\begin{align}\ce{ 2 H+ + 2e- &<=> H2 \\ 2 H2O + 2 e- &<=> 2 OH- + H2 \\ }\...


3

In water both $\ce{Cr^{3+}}$ and $\ce{Fe^{3+}}$ are in octahedral configurations. This means that $d$-orbitals become unequal in their energy; specifically 3 of them are lower and 2 are higher. This is the premise of 'crystal field theory'. Depending on the nearest neighborhood, the splitting may be strong enough to force electron pairing or it may not. For ...


3

The solution is to immerse each rod in both of the solutions (separately of course). Let's imagine that $\ce{A}$ is the more reducing metal. We can consider what happens in each case: If we immerse $\ce{A}$ in a solution of $\ce{A^{m+}}$ nothing will happen. Similarly, if we immerse $\ce{B}$ in a solution of $\ce{B^{n+}}$ nothing will happen. If we immerse $...


3

To construct reduction equations there are a set of steps you can follow. For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation. Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s)}$$ Under acidic ...


3

Natural variables of $\Delta G$ are $p$, $T$, and $N_i$. $p$ = pressure $T$ = Temperature $N_i$ = number of particles (or number of moles) The derivation showing this is listed on the Wikipedia.


3

Question 1: You are mixing the sign conventions for work from physics and chemistry. Sadly, they are not the same. In chemistry, the work done by the system on the surroundings is negative to signify energy leaving the system. Any transfer of energy (heat, work, etc.) from the system to the surroundings is denoted with a negative sign to ensure the energy ...


3

Basically, Imagine that you have a waterfall that is 100 meters high, and 50 meters long, this scenario makes reduction potential an intensive property: now let's say that you have two of the said waterfall--you will have in total (when put side by side) a larger waterfall that is 100 meters high, and 100 meters long. You do not need to multiply it by two ...


3

Easy is a big word. Lower potential means it requires less energy, but that doesn't say anything about how quick the reaction will proceed. A low potential reaction may still have slow kinetics and therefore not be "easy". If you are comparing ions as to which will produce more product as a result of an electrochemical reaction, you need to find ...


3

There are two ways to understand this equation. One is to realise that (reversible ideal case)$\Delta G=W_{\text{non-expansion}}$. Therefore, in an ideal chemical cell, if the potential difference between the electrodes is $E$, to move one mole electrons across the external circuit will be $FE$, which must be equal to the decrease in gibbs free energy of ...


3

There is a misunderstanding in the analysis of your first two half-reactions. Your first two half-reactions are fine. Remember that positive values of $E$ mean the reaction is spontaneous. Negative values of $E$ mean the reaction is nonspontaneous, since $$\Delta_\mathrm{r}G = -nFE_\mathrm{cell}.$$ Your first half-reaction is an oxidation (as written). ...


3

When you dip the electrode into the solution, it is indeed possible that electrolyte may interact with the surface of the electrode and there may be a limited amount of "charge flow" in the form of the creation of surface bound species. Without a bias created with the addition of an other electrode, however this reaction will be self terminating and it would ...


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