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Yes, you still use the Nernst equation, but one of your implicit assumptions is not correct. A point about the Nernst equation that often confuses people is that, at first glance, it doesn't predict any change in $E$ as temperature changes, as long as $Q$ remains equal to $1$, and $\ln Q = 0$. $$E = E^\circ - \frac{RT}{\nu F}\ln Q \tag{1}$$ However, a ...


5

The Nernst equation does not need a replacement for low concentrations. Yes the limit is $\infty$, which essentially means you can't realistically have zero concentration of anything. We may think of absolutely pure water as an idealized situation. Well, as soon as it gets in touch with copper, some $\ce{Cu^2+}$ ions instantly jump into the solution, and ...


4

The Nernst equation is as follows, \begin{align*} E_\text{red}&=E_\text{red}^{0}-\frac{RT}{nF}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}. \end{align*} where $c$ is the concentration of the species specified (usually the chemical activity, $a$, is used, but the equations follow the same principle). In the first case, $n=2$. So, \begin{align*}...


4

Just as you are supposed to use activity instead of concentration for dissolved components when applying Nernst equation, you are also supposed to use fugacity instead of the partial pressure for the gaseous components. The reason for that will become transparent if you follow the derivation of Nernst equation using chemical potential. Fugacity is defined ...


4

The simple mental test for the property being intensive or extensive is, what happens if you make the all the system twice as big. If the property remains the same, like concentration, or the cell electromotoric force ( voltage ), it is intensive. If you use the double volume of the solution, its concentration is still the same. The open voltage of Li-ion ...


3

Yes, it is possible to construct two half cells with the same standard reduction potentials. For example, we can half cells made of copper dipping in copper sulfate solution. The caveat is that if both the cells are under standard conditions, the resulting electrode potential would be zero. The trick is to use different concentration of copper ions in each ...


3

M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables: $R$ is a constant: $\pu{8.314 J mol-1 K-1}$ $F$ is also a ...


2

The 'Q' in Nerst equation is the same Q which you might've used while dealing with chemical equilibria, otherwise known as the Reaction Quotient. The expression is same as the equilibrium constant (K) and yes, they both include partial pressures and concentrations. The units of partial pressure is the same as the unit of pressure (atm or equivalent), which ...


2

For (1), I think you made the mistake of using a different value for overall cell potential when you worked backwards. Because if you use the correct overall cell potential, you get $K = 31.52$, your $\log K$ would be different if you used $0.685$ (which would make $\log K = 69.43$). Using the correct $E_\mathrm{cell}$ will get you the correct value of $n$ ...


2

Zhe's answer is the best. Though, I will try to offer a different (mathematical) approach, which will hopefully enable you to see the answer yourself, or rather, verify that your book's solution is indeed correct. According to you, the standard oxidation potential of magnesium electrode would be $E_{Mg/Mg^{2+}}=E^{º}_{Mg/Mg^{2+}}-\dfrac{0.059}{n}\,\mathbf{\...


2

The equation that corresponds to the standard reduction potential is: $$\ce{Mg^{2+} + 2e- -> Mg}$$ Notice that this is not the actual half reaction in your total equation, where magnesium is oxidized. But the book is determining the reduction potential, so the magnesium ion is a reactant not a product. In the reaction quotient, the concentration/...


2

(a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $q$ from one location with potential $\theta_1$ to one with $\theta_2$ is $q(\theta_2-\theta_1)$. This expression has the form of a constant factor ($q$) times the change in potential. In a gravitational field the potential ...


2

The Sackur-Tetrode equation can be written as $$S^{Trans}=R\ln \left( \frac{e}{C\Lambda^3} \right)$$ where $C$ is the concentration and $\Lambda$ the thermal wavelength. You can rewrite this as $$S^{Trans}=R\ln \left( \frac{e}{\Lambda^3 C^\circ\frac{C}{C^\circ}} \right) = R\ln \left( \frac{e}{C^\circ\Lambda^3} \right)-R\ln\left( \frac{C}{C^\circ} \right) = S^...


2

It is an interesting question. Daniel Harris is revising his book with my former mentor. Hope he clarifies this section in the revised version. Your point number 1 is misleading. The reason is that before the titration, theoretically there is no Fe(III). So Nernst equation should not be used- electrode potential is infinite (log 0 is undefined). When you ...


2

The Nernst equation for the anode is :$$E_a = E°_a + \frac{RT}{F}ln{\frac{[Li^+]_e [C_6]_s}{[LiC_6]_s}}$$ where the concentrations $[C_6]_s$ and $[LiC_6]_s$ are defined in the solid phase of the anode, and not in the electrolyte. The concentration $[Li^+]_e$ is defined in the electrolyte. When the anode is working, the concentration $[LiC_6]_s$ decreases in ...


2

You should read the article "Can One Take the Logarithm or the Sine of a Dimensioned Quantity or a Unit? Dimensional Analysis Involving Transcendental Functions" Here The concepts mentioned there will help for life even if you don't pursue chemistry. In short, natural log factor ($\ln$) should be dimensionless. Now your equation is $E=E^{\circ}+\...


1

Most thermodynamic equations are derived assuming a perfect gas or ideal solution. However, for real a solution the chemical potential $\mu_i=\mu^0_i +RT\ln(x_i)$ may not be accurate for the solvent where $\mu^0_i$ is the chemical potential of the liquid solvent at 1 atm and $x_i$ is the mole fraction. The solute may, similarly, not follow ideal behaviour. ...


1

problem where one is supposed to calculate the concentration of Fe3+ ions after a solution containing Fe2+ was titrated using acidified permanganate, There are several things which may be problematic in your approach. For the stated problem you do not need to invoke Nernst equation. Simply use the balanced equation. When potentiometric titration curve ...


1

Since battery scientists got Nobel this week, this question is worth pondering. It is not a trivial question. However consider this question: A car battery has 12 V, whereas 9 V cells are also common. If we connect two 1.5 V cells and one 9 V cell on series, one can generate 1.5+1.5+9 =12 V, yet this arrangement cannot start a car despite producing the same ...


1

As already said in the comments, you cannot use any unit you want in the reaction quotient; it has to be formulated in terms of activity (for solutes) or fugacity (for gases). The "standard state" of a chemical species is a reference state in which the properties of the chemical species are defined. This standard state is a theoretical state, it ...


1

A correct standard notation for the galvanic cell would also include the physical state of each species involved. As such, the correct notation for the given cell is: $$\ce{Cu(s)}~|~\ce{CuSO4(aq)}~(10^{-3}~\mathrm{M})~||~\ce{CH3COONa~(aq)}~(10^{-2}~\mathrm{M})~|~\ce{Pt(s)}~(\ce{H2(g)})~(1~ \mathrm{atm})$$ Recall that aqueous solutions of salts are ...


1

For a metal electrode immersed in a solution of its own salt, the concentration of the reduced form as a solid phase remains constant ($[\ce{M^0}] = \mathrm{const}$), therefore: $$E = E_0 + \frac{RT}{nF}\ln[\ce{Ox}]$$


1

This is why for equilibrium expressions we use activity and fugacity. Both of these values are dimensionless and are generally quite closely related to the concentration and partial pressure, respectively, for reasonable concentrations.


1

Most ions don't pass freely through the membrane. This is prevented by the hydrophobic (water "fearing") fatty acid tails of the phosholipids that provide the basis of the membrane. The regions outside and inside the membrane are aqueous and attract ions much more than does the hydrophobic fatty region in the interior of the membrane bilayer. Under usual ...


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