30

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 \...


20

Unfortunately, the question as stated is thermodynamically impossible. Let's look at the proposed reaction: $$\ce{CO2(g) -> CO(g) + O(g)}$$ This reaction is simply a bond dissociation (specifically, a carbon-oxygen covalent double bond is broken). We can look up the enthalpy change associated with it. From a table of values on Wikipedia, we find in the ...


19

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 ...


13

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

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 ...


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,...


9

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 (...


9

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- &<...


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

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

You’re looking at bond dissociation energies. They, however, do not give a good picture. A better place to start looking is the standard enthalpy of formation. The linked Wikipedia article provides an extensive list of compounds but only two matter: $\displaystyle\Delta_\mathrm fH^0 (\ce{CO}) = \pu{-110.525 kJ/mol}$ $\displaystyle\Delta_\mathrm fH^0 (\ce{...


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

Besides the the thermodynamic aspects discussed by Matthew, consider the kinetic aspects. Oxidation by a chlorine oxyanion involves displacement 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 ...


7

A salt like $\ce{Na2SO4}$ is essential in electrolysis. It provides ions $\ce{Na+}$ and $\ce{SO4^{2-}}$ which are attracted by the electrodes in the solution and migrate to them. When they arrive near the electrodes, they are not discharged. But they neutralize the charges of the ions that are produced out of water being destroyed at these electrodes. Let's ...


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

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

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, ...


6

You may heat mixture of $\ce{MnO2}$ with charcoal to higher than $\pu{850 °C},$ in the absence of air. Manganese will be produced in the solid state by reduction of the oxide. But it is sensitive to air oxidation and easily reoxidized. So better use an excess of charcoal. In the industry, rough $\ce{MnO2}$ out of the mine is first reduced to $\ce{MnO}$ by $\...


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

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 ...


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

It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation. A generic Latimer diagram $$\ce{A ->[$E_1$] B ->[$E_2$] C}$$ posesses the following properties: If $E_2 > E_1,$ then $\ce{B}$ is thermodynamically unstable and ...


4

You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions. Standard thermodynamic conditions are usually A temperature of 298 K A pressure of gaseous components of 1 atm (or 1 bar) A concentration of 1 M The reference reduction potentials you quote are all standard potentials. ...


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 ...


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