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You could try considering the Cottrell equation or the mass transfer limiting case of the butler-volmer equation $$ i = \frac {nFAc_{j}^{0}\sqrt{D_{j}}}{\sqrt{\pi t}} $$ i = current, in unit A n = number of electrons (to reduce/oxidize one molecule of analyte j, for example) F = Faraday constant, 96485 C/mol A = area of the (planar) electrode in cm2 ...


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$$\ce{Mn^{2+} + Cr2O7^{2-} -> Mn^{3+} + Cr^{3+}}$$ If you were to balance this reaction, using the appropriate molecules of water and protons, you'd get: $$\ce{ Mn^{2+} + Cr2O7^{2-} + 14H+ + 5e- -> Mn^{3+} + 2Cr^{3+} + 7H2O }$$ The presence of protons the left side of equation suggests that this reaction has a higher rate when the concentration of ...


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Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied. However, there is no requirement that a path between two states ...


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A battery has a small shelf life if it discharges internally. Saying that a discharged battery is in chemical equillibrium (or much closer to it than a fresh one) is a true statement, obviously, but nothing else. If you reverse the polarity of a primary battery, you get some other reaction than the one happening during discharging backwards. That says ...


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You can imagine the cell electrodes as capacitors, that are charged by chemical way. But the capacitance of these capacitors $$C=\frac {\mathrm{d}q}{\mathrm{d}E}$$ is very small, so does the accumulated charge $q$. The charge loses due pushing electrons through the wires of a closed circuit are balanced by simultaneous recharging by ongoing redox ...


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is it true that if I connect an aqueous oxidant to an aqueous reductant via a wire, will a current flow along the wire? In particular, does current flow in the following scenario? No true. Nothing will happen as the circuit is incomplete. The oxidant or the reductant species have to see each other "face to face". Guess what is the most powerful reducing ...


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Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction. Step 1: Start with the given metal species, with the metal oxidation states included: $\ce{M^VO4^{3-}->M^{III}O(OH)}$ Step 2: Add the electrons according to the oxidation states. Here the oxidation state on the metal drops by two, so there ...


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The explanation given to you is rather incorrect. Dilute sulfuric acid is added to iron(II) or even iron(III) salts is added to prevent hydrolysis of the salts. What happens when you don't add an acid to iron(II) solution in water. With time, you will see that a brown precipitate is forming and settling to the bottom. Iron forms hydroxides which are very ...


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You got oxidation numbers mixed up. As the comments pointed out, the +9 state is possible only in very few elements and mercury is not one of those. To my understanding in general chemistry, this problem can easily solved using Solubility Rules: Salts containing Group I elements ($\ce{Li+, Na+, K+, Cs+, Rb+}$) and salts containing ammonium ion ($\ce{...


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You got the oxidation numbers in the $\ce{NO3^-}$ anion wrong. Remember that the sum of the oxidation numbers in a species must be equal to the total charge. You claimed oxygen has a $-2$ oxidation number and nitrogen has $-3$, this would mean that you would not have an $\ce{NO3^-}$ ion but an $\ce{NO3^9-}$, which is clearly not the case. Try fixing the ...


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You cannot do simple volumetric redox titration to determine iron content because the amount of iron is very very small in a single fruit (on the order of fraction of a milligram). Imagine what would be the buret reading? Classical methods are good for large concentrations >> 1% wt/wt With such small quantities, UV-Vis absorption spectroscopy or atomic ...


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It is rather: $$\ce{2 Al + 6 H2O + 2 OH- -> 2 [Al(OH)4]^- + 3 H2 ^}$$where $\ce{OH-}$ comes either from hydroxide dissociation, either from carbonate hydrolysis. $$\ce{CO3^2- + H2O <=> HCO3- + OH-}$$ The reaction with carbonate would gradually slow down as the carbonate/bicarbonate buffer will kick in. Initial $\mathrm{pH}$ is ( see notes below ...


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Well I am just a high school student, so I may be wrong, and I beg pardon in advance, but I would like to share my notions about this topic. @ChakravarthyKalyan I would like to ask you to review your last step in the conversion of 3d to 1. 3d on oxidation should yield this product (in the image)--which is by my knowledge not same as 1. So even your ...


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