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6

I get that this makes sense, but I have no idea how to derive it myself. Any help would be appreciated! Good question because such questions were pursued by prominent physical chemists and Nobel laureates (Ostwald/ Nernst and Hoffmann). The reduction of water can be tested visually and the apparatus to demonstrate it indeed named after Hoffman. Basically, ...


3

First of all: Reduction is just a fancy way of saying "gain of electrons". The posed question can thus be reformulated as: What happens to water, if it gains more electrons? In order to answer this, we should have a look at the oxidation numbers of water: Since oxygen is more electronegative, it has an oxidation number of -II, while the two ...


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 ...


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 ...


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 ...


2

Reduction processes as posed here, “rearranges” the atoms upon taking up two electrons. One way you may have been able to “derive” or take a good guess is to consider the electrolysis of water, breaking water down into $\ce{H2}$ and $\ce{O2}$. In this case there has to be both an Oxidation and a Reduction. In the hydrolysis case, the $\ce{H2}$ is the ...


2

Both the answers above are accurate and touch on my point here. There are two factors that contribute to whether a reaction is spontaneous (if it will happen on its own) such as the one you described (iron with oxygen). The thermodynamic properties of the reaction (like the net heat exchange, exothermic vs. endothermic, etc.) The entropy of the reaction. ...


1

Permanganate oxidizes iodide to iodine $\ce{I2}$ at all pH values. In acidic conditions, the reaction is finished here. But in basic solution, the iodine $\ce{I2}$ is transformed into iodide and iodate according to $$\ce{3I2 + 6 OH^- -> 5 I- + IO3^- + 3 H2O}$$The mixture iodide + iodate reacts the other way as soon as the solution becomes acidic : $$\ce{...


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) ...


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I'm not sure, but I think krypton difluoride is the strongest oxidizing agent I know of that's fairly stable at normal temperatures and pressures. (It decomposes at ~10% per hour at ~25°C, but is stable at dry ice temperature or below. Wikipedia claims it is "the most powerful known oxidizing agent known" and "more powerful even than elemental ...


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