8
Energy storage depends on the electromotive potential (i.e. difference between species in the electromotive series) and on the number of electrons available.
Li, for example, has an oxidation potential of ~3.04 V relative to hydrogen, but Al has one of 1.66 V, so Li has the greater potential. On the other hand, Li has only one freely available outer ...
6
Let's write the half-cell equations in standard form as half-cell reductions.
In a table of standard electrode potentials it would silly to write both the reduction and oxidation reactions since that would needlessly double the size of the table.
$$
\begin{align}
\ce{A+ + e- &-> A} &\quad V_\ce{A} \\
\ce{B^{2+} + 2e- &-> B} &\quad ...
4
I have done a significant amount of research over the past ten years to trace to origins of these electrochemical conventions and luckily got a chance to discuss these with some top electrochemists. I have been planning to write an article on this issue since it is a perpetual confusion. Basically, the origin of these "signs" issues originated in Germany and ...
4
According to my notes and many sources on the internet, electrons and cations both travel from the anode (A in the image) to the cathode (B in the image).
The idea of the salt bridge is to prevent electrolytes mixing while providing ion flow. When you have a high concentration of inert ions in the salt bridge, cations in the salt bridge will flow into B, ...
4
Note that the actual potential for a particular redox reaction is not a fixed value, but depends on concentrations ( more exactly activities ) of reagents. The standard redox potentials are potentials with activities equal to 1,
If we consider reactions
$$\begin{align}\ce{
2 H+ + 2e- &<=> H2 \\
2 H2O + 2 e- &<=> 2 OH- + H2 \\
}\...
3
I still teach switching the sign. I find it easier to remember adding up reduction potential and oxidation potential. The half reactions are written as reduction in a table of reduction potentials, so it makes sense that you have to treat the oxidation half reaction differently.
If the cell potential is calculated from reduction potential of the cathode ...
3
The main thing to point out is that your question doesn't actually make sense the way you want it to.
What does it mean if a reduction potential is zero? Or positive? Or negative? Is that favorable or not? It turns out we don't specify this at all!
As an example, consider the standard hydrogen electrode and it's reduction potential:
$$\ce{2H+ + 2e- -> ...
3
To the extent that it's appropriate to treat it as a thermodynamic quantity, given that it's generally a description of a specific, macroscopic system (and thus violates the statistical-mechanical large-numbers assumption of thermo), resistance is extensive, always.
Resistance of a wire: The resistance of a wire depends on area and length. This implies ...
2
First off, read carefully:
Different metals will increase the likelihood of corrosion.
It doesn't say:
Different metals will increase the likelihood of corrosion of gold
With that in mind, it's called galvanic corrosion and occurs when contact between two (or more) dissimilar metals causes the more reactive one to corrode when it normally would not ...
2
Of course, we're starting by acknowledging a perpetual motion machine is impossible. The question, then, is how do we understand, through chemical thermodynamics, why your specific set of steps can't constitute a perpetual motion machine.
The answer is that the electrical energy required for your step 2 is much greater than you get back from your step 4, ...
2
Free energy is a state function. No matter how you run the reaction, if you start with a certain state (set of concentration) and end with a certain state (set of concentrations at equilibrium), the change will be the same.
The free energy of reaction is the maximal amount of non-pV work (electrical work in this case) the reaction can do. If it does not do ...
2
My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.
That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction ...
2
You're questioning the intuitive disconnect caused by most galvanic cell drawings which seem to assume the electrolyte solution in the salt bridge does not conduct electricity, so let's investigate.
Imagine a Zn/Cu$^{2+}$ cell with electrodes 5 cm apart in a 3.5% NaCl solution with a tube (1 cm$^2$ cross-section) of solution as the salt bridge for balancing ...
2
My book tells me to keep the E∘half-cells as they are written in the
tables and simply put them in
Your book is then one of the few books which teaches electrochemistry properly. The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to hydrogen electrode. I don;t know if ...
2
Take a look at the two half reactions:
$$
\begin{align}
\ce{Ag+(aq) + e- &→ Ag(s)} &\qquad E^\circ &= \pu{0.80 V} \\
\ce{Sn^2+(aq) + 2 e- &→ Sn(s)} &\qquad E^\circ &= \pu{-0.14 V}
\end{align}
$$
If there is an electron for grabs (like the ones in the wire of a voltaic cell), $\ce{Ag+(aq)}$ and $\ce{Sn^2+(aq)}$ are competing for it. ...
2
The above post answers your questions, but let me add that
$$\ce{2 H+ + 2 e- <=> H2}\label{rxn:1}\tag{1}$$
and
$$\ce{2 H2O + 2 e- <=> H2 + 2 OH-}\label{rxn:2}\tag{2}$$
are two different experimental procedures. Hence these are two distinct half cell reactions. It happens that you get $\ce{H2}$ as a product in $\eqref{rxn:1}$ and $\eqref{rxn:...
2
It is a bad question because "Metal A is more reactive than Metal B" has no real meaning. It is a vague choice of word, A is more reactive, but with respect to what?
Your logic is correct
Since A is more reactive, it undergoes oxidation to form A2+ ions,
while B2+ ions undergo reduction and form B. From this we can deduce
that A is the anode and B is ...
2
Equilibrium constant
When a reaction is at equilibrium, the species will have concentrations called the equilibrium concentrations. They do not depend on whether the reaction is written in one direction or in the reverse direction, and whether all the stoichiometric coefficients are multiplied by the same value. However, your equilibrium expression will be ...
2
As noted in your question, the standard reduction potentials are given as such in tables.
$$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$
...
1
The sign of the electrode reduction potential is invariant. If reflects the sign of the electrostatic charge of the electrode with respect to the hydrogen electrode.* Also remember that all electrode potentials are written as reduction these days. This is a convention set by all electrochemists all over the world along with the other conventions. The ...
1
You do need the electromotive series or reactivity series to answer that question. If you will not have access to it during such exams, you'll need to memorize the order of some common elements, and to understand the arrangement of the Periodic Table.
The most reactive metals are at the bottom left of the table, low in Group I. The "noble" or coinage metals ...
1
It is a bit of a tricky question. The trick is in the wording "flow through". The positive ions on the A side do not actually "flow through" the salt bridge from A to B. Rather positive ions exit the salt bridge on the B side to compensate for the cations depositing on the cathode (B). Likewise negative ions exit the salt bridge on the A side to balance the ...
1
A cell with two electrodes separated from each other and an aluminum divider plate in the middle is really two cells: one has an aluminum plate as a cathode, and the other cell has the same aluminum plate as the anode.
In this situation, the solution does not directly connect the external anode and cathode; the aluminum divider separates the solution into ...
1
The Nernst equation and electrochemical potentials relate to redox systems, not to reagents and products. The forward and reversed reactions are the same redox system.
Imagine you would want to make a galvanical cell with the same electrodes. Flipping the sign would grant you a Nobel price for inventing a perpetuum mobile.
1
My question is that suppose we have to metals, A and B. A produces +1
ions. B produces +2 ions. Now even if the two have the same tendency
to lose or gain electrons, the B plate would be 'more negative'( As in
this plate would have more negative charge). How has the voltage
helped us determine the difference in the tendencies of these two
metals?
...
1
We should not extend the idea of electrode potentials for very high concentrations such as concentrated sulfuric acid which is on the order of 18 M. The interesting property of these acids (HNO3 or H2SO4) is that they behave as oxidizing agents or even as dehydrating agents (H2SO4) especially at high temperature and at high concentrations. This is not to be ...
1
All depends on the cell chemistry and geometry. It is usually a combination of keeping the energy and conversion of the free energy to thermal energy, when reagents diffuse and happen to meet each other. Or, some side reaction with solvent or auxiliary components may occur, like for $\ce{Li-ion}$ cells.
Some primary lithium cells last many years, while some ...
1
The anode is the electrode, where substances are losing electrons and are oxidated.
The cathode is the electrode, where substances are gaining electrons and are reduced.
The tricky part for the memorising is, anodes and cathodes flip the position, when the current is reversed, depending on if
the cell is in the mode of electrolysis
the cell is in the mode ...
1
There is no completed electronic circuit in an electrochemical cell
In an electrochemical cell, the anode is the source of electrons to the external circuit and the cathode is the sink. The circuit of charge transport gets completed by ions traveling inside the cell. A solar cell is different from an electrochemical cell in that their is no net chemical ...
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