A voltaic cell requires a positive voltage which indicates a spontaneous reaction.
The correct answer is positive.
To form a voltaic cell, the voltage should be positive. In a voltaic cell, electricity is produced as a result of a redox reaction.
This is determined by the reduction potentials of the half reactions. If the cell potential (voltage) is positive, this indicates a spontaneous reaction, meaning it can occur naturally without outside energy input.
The voltaic cell consists of two half cells, each containing a half reaction. The half cell with the oxidation reaction is the anode, considered the negative electrode of the cell.
Conversely, the cathode, containing the reduction reaction, is the positive electrode of the cell. Even though the anode is negative and the cathode is positive, electrons flow from the anode to the cathode in a voltaic cell.
Make sure you understand these concepts well as they are frequently used in batteries, a very practical application of voltaic cells.
A positive cell voltage verifies a spontaneous reaction, whereas the presence of a negative cell potential indicates a non-spontaneous reaction. In the latter case, to make the reaction occur, a voltage greater than the cell potential must be introduced to the cell, forcing the reaction to proceed.
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Enzyme-catalyzed Hydrolysis: Figure 8 illustrates a simple mechanism enzyme-catalyzed hydrolysis. The mode of action for the emulsin enzyme is the Michaelis-Menten mechanism (one you will study in BIOC*2580). The substrate salicin [S] attaches itself at the active site of the emulsin enzyme [E] to form an enzyme-substrate complex [ES]. The [ES] complex dissociates to produce the product species
[P]n (n = number of products) and the original enzyme [E].
With the reaction taking place in an aqueous environment, the amount of water consumed is negligible compared with the amount of water originally present, so its concentration remains constant. As before, the enzyme is regenerated and so its concentration is also constant. The rate law expression can again be simplified because both concentration values can be absorbed into a modified rate constant, e.g. k', where k'= k[E][H20].
In addition, this particular experiment has the salicin in excess of the enzyme emulsin.
As a result, the enzyme is effectively saturated with the substrate. This means the addition of more substrate will not accelerate the reaction. Under these experimental conditions, virtually all the enzyme is present as an enzyme-substrate complex and the rate is independent of the salicin concentration and will now obey zero-order kinetics. The observed rate law, based on our experimental design, becomes
Rate = Kobserved Where Kobserved = k [SIE][H20].
After evaluating the mechanism provided in Figure 8 to determine the theoretical rate law for the enzyme-catalyzed reaction, what was the overall order?
Question 6 options:
a)
0
b)
1
c)
2
d)
3
e)
None of the above.