25

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 \...


18

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

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 ...


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 ...


8

The dichromate ($\ce{Cr2O7^2-}$) ions are strong oxidizing agents at low pH. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. In redox reactions in aqueous acid solution, the aquated $\ce{Cr^3+}$ ion is produced according to following half-reaction (...


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 (...


7

First, hypochlorite ($\ce{ClO−}$) is not a strong oxidizing agent compared to other chloride oxyanions, at least in acidic medium. Under standard conditions in acidic medium, chlorous acid ($\ce{HClO2}$) is the best oxidizing agent among them due to the largest positive number for standard electrode potential: $$\begin{align} \ce{HClO2 + 2H+ + 2 e- &<...


6

I'm surprised your textbook did not derive this equation from the reaction isotherm relationship between $\Delta G$ and the reaction quotient $Q$ and the Nernst equation. The derivation is not hard. Reaction isotherm equation: $$\Delta_\mathrm{r} G =\Delta_\mathrm{r} G^\circ + RT\ln Q$$ Nernst equation: $$E_\mathrm{cell} = E^\circ_\mathrm{cell} - \frac{...


6

There are two ways to understand this equation. One is to realise that (reversible ideal case) $\Delta G = W_\text{non-exp}$ (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 ...


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, helium, ...


5

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 ...


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 ...


5

According to the definition used by IUPAC, the standard electrode potential $E^\circ$ of the standard hydrogen electrode is zero at all temperatures. For solutions in protic solvents, the universal reference electrode for which, under standard conditions, the standard electrode potential ($\ce{H+}/\ce{H2}$) is zero at all temperatures. The absolute ...


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

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

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 ...


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

Electrodes transport electrons produced from one half-cell to another, thus producing an electric charge. You need to have a reference value to compare it to in order to obtain the relative potential. Relative to the Standard Hydrogen Electrode (SHE), that is. The SHE has a potential that is set arbitrarily to zero giving you a basis with which to compare ...


4

Potassium dichromate is a strong oxidizing agent and it helps any other compound to oxidize by itself getting reduced to $\ce{Cr^3+}$. This is a general redox reaction. Normally, the counterion i.e the anion comes from the acid used. If $\ce{H2SO4}$ is used then $\ce{Cr2(SO4)3}$ will form; if $\ce{HCl}$ is used, then $\ce{CrCl3}$ will form. Take for example: ...


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 \\ }\...


4

Besides the the thermodynamic aspects discussed by Matthew, consider the kinetic aspects. Oxidation by a chlorine oxyanion involves displacenent of oxygen from its bond with the chlorine. Such a displacement, in a protic solvent such as water, must involve protonation of the oxygen: Oxygen without the proton would have to be displaced as oxide ion, which ...


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 $\ce{...


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