Calculate for the electrochemical cell below,
Ag(s) | AgCl(s) | Cl–(aq) || Cu2+(aq) | Cu(s)
given the following standard reduction potentials.
Cu2+(aq) + 2 e– -> Cu(s) E = +0.337 V
AgCl(s) + e– -> Ag(s) + Cl–(aq) E = +0.222 V
a. –0.115 V
b. –0.107 V
c. +0.115 V
d. +0.452 V
e. +0.559 V
Answer A
I think when the cell is set up as written above (with the diagram) it means
Ag + Cl^- ==> AgCl + e E = -0.222
Cu^2+ + 2e --> Cu E = 0.337
So the sum is +0.115 or c.
Well, let me calculate it for you in my own clowny way:
Okay, let's think about this circus act of a electrochemical cell. We have a silver clown (Ag) in the left side and a copper clown (Cu) in the right side. Between them, we have this salty clown called AgCl and its partner in crime, Cl⁻.
Now, according to the standard reduction potentials provided, the copper clown (Cu²⁺ + 2e⁻ -> Cu) has an E° of +0.337 V, while the silver clown (AgCl + e⁻ -> Ag + Cl⁻) has an E° of +0.222 V.
Since we're dealing with a galvanic (or spontaneous) cell, the overall cell potential can be calculated by subtracting the larger reduction potential from the smaller reduction potential.
In this case, we have +0.222 V (silver clown) - +0.337 V (copper clown) = -0.115 V.
So, the answer is a. -0.115 V.
Remember, life is like an electrochemical cell, full of positive and negative charges. Keep clownin' around!
To calculate the cell potential (Ecell) for the given electrochemical cell, we need to determine the reduction half-reactions occurring at the cathode and anode.
Given half-reactions:
1. Cu2+(aq) + 2e– → Cu(s) (E° = +0.337V) --> Cathode half-reaction
2. AgCl(s) + e– → Ag(s) + Cl–(aq) (E° = +0.222V) --> Anode half-reaction
To calculate Ecell, we can subtract the reduction potential of the anode half-reaction (AgCl) from the reduction potential of the cathode half-reaction (Cu2+):
Ecell = E°cathode - E°anode
Ecell = (+0.337V) - (+0.222V)
Ecell = +0.115V
Thus, the correct answer is c. +0.115V.
To solve this question, we can use the Nernst equation, which relates the standard reduction potentials to the cell potential under non-standard conditions:
E = E° - (0.0592/n) * log(Q)
where:
E = cell potential under non-standard conditions
E° = standard cell potential
0.0592 = a constant at room temperature
n = number of electrons transferred in the balanced equation
Q = reaction quotient
First, let's write the balanced equation for the cell:
Cu2+(aq) + 2 e– -> Cu(s)
AgCl(s) + e– -> Ag(s) + Cl–(aq)
The overall equation for the cell can be obtained by adding these two half-reactions:
Cu2+(aq) + 2 AgCl(s) -> Cu(s) + 2 Ag(s) + 2 Cl–(aq)
Now, let's calculate the reaction quotient Q. The concentrations of the species involved in the reaction quotient are as follows:
[Cu2+] = concentration of Cu2+ (unknown)
[AgCl] = concentration of AgCl (unknown)
[Ag+] = concentration of Ag+ (unknown)
[Cl–] = concentration of Cl– (unknown)
Since the question doesn't provide concentrations, we can assume they are all 1.
Therefore, Q = [Ag+]^2 * [Cl–]^2 / ([Cu2+] * [AgCl]^2)
Now, let's substitute the given values into the Nernst equation:
E = E° - (0.0592/n) * log(Q)
E = 0.337 V - (0.0592/2) * log(1^2 * 1^2 / (1 * 1^2))
E = 0.337 V - (0.0296) * log(1)
E = 0.337 V - (0.0296) * 0
E = 0.337 V - 0
E = 0.337 V
The cell potential under non-standard conditions is 0.337 V.
Comparing this value with the given options, none of the provided options match. Therefore, the correct answer is not amongst the given options.