The potential of hydrogen electrode is -118MV.thr H^+ concentration of the solution is
a)0.01M
b)2M
C)10^4M
d)1M
Taking E =118millivolt...
2H+. +2e ...H2
E= E*-0.059/2log H2/[H+]2
-118=0-0.059/2log 1/[H+]2
Solving it by using antilog we will get 0.01
To determine the pH of a solution using the potential of a hydrogen electrode, we can use the Nernst equation:
E = E° - (RT / nF) * ln([H+])
Where:
- E is the potential of the hydrogen electrode
- E° is the standard potential of the hydrogen electrode (which is 0V)
- R is the gas constant (8.314 J/(mol*K))
- T is the temperature in Kelvin
- n is the number of electrons transferred in the oxidation/reduction reaction (for hydrogen electrode, it is 1)
- F is the Faraday's constant (96485 C/mol)
- [H+] is the concentration of H+ ions in the solution
Using the given potential of -118mV (which is -0.118V), we can substitute the values into the Nernst equation:
-0.118 = 0 - (8.314 * T / (1 * 96485)) * ln([H+])
Since [H+] = 10^-pH, we can rewrite the equation as:
-0.118 = - (8.314 * T / (1 * 96485)) * ln(10^-pH)
0.118 = (8.314 * T / (1 * 96485)) * pH * ln(10)
Simplifying further, we get:
pH = (0.118 * 96485) / (8.314 * T * ln(10))
To determine the pH for each given concentration, we need the value of T (temperature). Assuming T is room temperature (~ 298 K), we can calculate the pH for each option:
a) [H+] = 0.01 M
pH = (0.118 * 96485) / (8.314 * 298 * ln(10))
= 0.040
b) [H+] = 2 M
pH = (0.118 * 96485) / (8.314 * 298 * ln(10))
= -0.697
(Note: The calculated pH is negative, which is not possible for a solution with a concentration above 1 M. Therefore, this option is not valid.)
c) [H+] = 10^4 M
pH = (0.118 * 96485) / (8.314 * 298 * ln(10^4))
= 7.000
d) [H+] = 1 M
pH = (0.118 * 96485) / (8.314 * 298 * ln(10))
= 0.000
Therefore, the correct answer is:
a) 0.01 M, pH = 0.040
To determine the hydrogen ion concentration of the solution, we can use the Nernst equation, which relates the potential of a half-cell to the concentration of the species involved.
The Nernst equation for a hydrogen electrode is given by:
E = E° + (0.0592/n) * log([H+])
Where:
- E is the potential of the hydrogen electrode,
- E° is the standard electrode potential of the hydrogen electrode (0V),
- n is the number of electrons involved in the reaction (2 for the hydrogen electrode), and
- [H+] is the concentration of hydrogen ions.
Given that the potential of the hydrogen electrode is -118 mV (-0.118 V), we can substitute these values into the Nernst equation and solve for [H+] to find the concentration of hydrogen ions.
For option a) 0.01 M:
E = 0 + (0.0592/2) * log(0.01)
E = 0 + 0.0296 * (-2)
E = -0.0592 V
The potential of -0.0592 V does not match the given potential of -0.118 V, so the hydrogen ion concentration is not 0.01 M.
For option b) 2 M:
E = 0 + (0.0592/2) * log(2)
E = 0 + 0.0296 * (0.301)
E = 0.0089 V
The potential of 0.0089 V does not match the given potential of -0.118 V, so the hydrogen ion concentration is not 2 M.
For option c) 10^4 M (10,000 M):
E = 0 + (0.0592/2) * log(10^4)
E = 0 + 0.0296 * (4)
E = 0.1184 V
The potential of 0.1184 V does not match the given potential of -0.118 V, so the hydrogen ion concentration is not 10^4 M.
For option d) 1 M:
E = 0 + (0.0592/2) * log(1)
E = 0 + 0.0296 * (0)
E = 0 V
The potential of 0 V matches the given potential of -0.118 V, so the hydrogen ion concentration is 1 M.
Therefore, the correct answer is d) 1 M.
I assume you mean -118 millivolts and that pH2 = 1 atmosphere..
-0.118 v = Eo - (0.06/n)log (H2/(H^+)^2 for the equation
2H^+ 2e ==> H2
Post your work if you get stuck.