An electrochemical cell of notation Pd | Pd²⁺ || Cu²⁺ | Cu has E° = -0.65 V. If we know that the standard reduction potential of Cu²⁺/Cu is E° = 0.34 V, what is the standard reduction potential for Pd²⁺/Pd?
Pd | Pd²⁺ || Cu²⁺ | Cu
.......Pd(s) ==> Pd^2+(aq) + 2e..........Eo = ?
......Cu^2+(aq) + 2e ==> Cu(s)............Eo = 0.34
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.....Pd(s) + Cu^2+(aq) --> Pd^2+(aq) + Cu(s) Ecell = -0.65
Ecell = ? + 0.34
-0.65 = ? + 0.34
? = -0.65 - 0.34 = -0.99
So Pd(s) ==> Pd^2+(aq) + 2e has Eo value of -0.99; however this is written as an oxidation. The problems wants the reduction potential; therefore, the redn pot is +0.99
Ah, let me put on my electrode hat and crunch some numbers for you. So, we have the overall cell notation as Pd | Pd²⁺ || Cu²⁺ | Cu, with an E° of -0.65 V for the complete cell. Now, we know that the standard reduction potential for the Cu²⁺/Cu half-cell is 0.34 V.
To find the standard reduction potential for Pd²⁺/Pd, we can use the equation:
E°(cell) = E°(cathode) - E°(anode)
Since the Pd half-cell is acting as the anode, we have:
E°(cell) = E°(Cu²⁺/Cu) - E°(Pd²⁺/Pd)
Plugging in the values we know:
-0.65 V = 0.34 V - E°(Pd²⁺/Pd)
Now it's just a matter of rearranging the equation to solve for E°(Pd²⁺/Pd):
E°(Pd²⁺/Pd) = 0.34 V - (-0.65 V)
E°(Pd²⁺/Pd) = 0.99 V
So, the standard reduction potential for Pd²⁺/Pd is a zesty 0.99 volts. Keep that in your electrifying repertoire, my friend!
To find the standard reduction potential for Pd²⁺/Pd, we can use the Nernst equation:
E = E° - (0.0592/n) * log(Q)
where E is the cell potential, E° is the standard reduction potential, n is the number of electrons transferred, and Q is the reaction quotient.
In this case, the reaction in the cell is:
Pd²⁺ + 2e⁻ → Pd
So, n = 2.
The given cell notation is Pd | Pd²⁺ || Cu²⁺ | Cu. According to the notation, the reactant in the left compartment (Pd | Pd²⁺) is the anode, and the reactant in the right compartment (Cu²⁺ | Cu) is the cathode.
The Nernst equation for the anode reaction is:
E(anode) = E°(anode) - (0.0592/2) * log(Q(anode))
The Nernst equation for the cathode reaction is:
E(cathode) = E°(cathode) - (0.0592/2) * log(Q(cathode))
Since the standard reduction potential for Cu²⁺/Cu is given as E° = 0.34 V, the cathode reaction can be written as:
Cu²⁺ + 2e⁻ → Cu
Using the Nernst equation for the cathode reaction:
E(cathode) = 0.34 - (0.0592/2) * log(Q(cathode))
By comparison, since the standard reduction potential for Pd²⁺/Pd is not given, let's assume it as E°(anode). Thus, for the anode reaction:
E(anode) = E°(anode) - (0.0592/2) * log(Q(anode))
Now, we can write the overall cell reaction by combining the anode and cathode reactions:
Pd²⁺ + Cu → Pd + Cu²⁺
The reaction quotient for the overall cell reaction is:
Q = [Pd][Cu²⁺] / [Pd²⁺][Cu]
Substituting the values into the Nernst equation for the overall cell:
-0.65 = E°(anode) - (0.0592/2) * log(Q)
Now, let's substitute the given standard reduction potential for Cu²⁺/Cu into the Nernst equation for the cathode reaction:
-0.65 = E°(anode) - (0.0592/2) * log(Q) - (0.0592/2) * log(1)
Since log(1) = 0, we can simplify the equation:
-0.65 = E°(anode) - (0.0592/2) * log(Q)
Now, let's isolate E°(anode):
E°(anode) = -0.65 + (0.0592/2) * log(Q)
Since the concentrations of Pd, Cu²⁺, and Pd²⁺ are not given, we cannot determine the exact value of the standard reduction potential for Pd²⁺/Pd. However, using this equation, you can calculate the value of E°(anode) given the concentrations of the ions involved in the reaction.
To find the standard reduction potential for Pd²⁺/Pd, you can use the equation:
E° cell = E° cathode - E° anode
In the given electrochemical cell notation, the reduction half-reaction is Pd²⁺ + 2e⁻ → Pd, while the oxidation half-reaction is Cu → Cu²⁺ + 2e⁻.
Using the given standard reduction potential for Cu²⁺/Cu as E° = 0.34 V, we can substitute these values into the equation:
E° cell = E° cathode - E° anode
-0.65 V = E° cathode - 0.34 V
Now, we can rearrange the equation to solve for E° cathode:
E° cathode = E° cell + E° anode
E° cathode = -0.65 V + 0.34 V
E° cathode = -0.31 V
Therefore, the standard reduction potential for Pd²⁺/Pd is -0.31 V.