[H3O+] = 3.6 x 10-2 M
1 answer:
Answer:
Explanation:
Hello there!
In this case, since this concentration of hydronium ions is given, we infer we may calculate that of the hydroxide ions, pH and pOH because no question was specifically given. That is why we proceed as follows:
- pH: Here, we use this concentration to calculate the pH according to the negative logarithm:
Which means that the solution is acidic as the pH is less than 7.
- pOH: since the pH added to the pOH is 14, we can calculate the latter as follows:
- Hydroxide concentration: Given the pOH, we apply the antilogarithm to calculate such concentration as shown below:
Keep in mind, this is an arbitrary solution because no question was provided.
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Answer : The value of and K is, -180 kJ/mol and
Explanation :
The balanced cell reaction will be,
The half-cell reactions are:
Oxidation reaction (anode) :
Reduction reaction (cathode) :
Relationship between standard Gibbs free energy and standard electrode potential follows:
where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:
Now we have to calculate the value of 'K'.
where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant =
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:
Therefore, the value of and K is, -180 kJ/mol and