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KCl is used as salt bridge because it provides positive K+ ions and negative Cl- ions as the salt bridge needs to maintain the neutrality in the system by providing enough negative ions equal to the positive ions during oxidation.
Hydrogen gas is not liberated when the following metal is added to dil. HCl.
The metals, present below hydrogen in the electrochemical series, cannot liberate hydrogen from the dilute acids.
Among the given metals only Ag is present below hydrogen in electrochemical series, so it does not evolve hydrogen with dil HCl.
Ag−I−dilHCl ⟶ No reaction
Temperature for the measurement of standard electrode potential is:
The standard electrode potentials are customarily determined at solute concentrations of 1 Molar, gas pressures of 1 atmosphere, and a standard temperature which is usually 25°C i.e, 298 K.
The reduction potential of an element A is 1.71 V. What can be concluded from this?
Stronger oxidizing agent means more easily it undergoes reduction and thus have greater standard reduction potential.
Remember, Li is the strongest reducing agent and F is the strongest oxidizing agent
Consider the following reaction which of the following statement is true for this cell reaction.
(Zn + Cu2+ → Zn2+ + Cu)
A half cell reaction A- → A + e- has a large negative reduction potential. It follows that :
The more negative the potential is the more likely the element will be oxidized. Hence, A− is readily oxidized.
Two elements A and B have reduction potential’s 0.23V and 0.46V which statement is true regarding these two elements.
The more positive the reduction potential value, the more is the tendency of the element to reduce itself and hence more will be the tendency to accept electrons.
Hence the other element will have more tendency to lose electrons
The reduction potential of an element A is -2.71V. What can be concluded from this?
Reduction potential means to accept electrons to reduce oneself.
A + e- → A- ∆Ereduction = +ve value
Since, the reduction potential is negative, it means that the reaction will reverse to make ∆E value +ve. So the reaction becomes,
A → A+ + e-
This becomes oxidation of A. So oxidation of A will be easy.
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