Consider the voltaic cell:
Zn(s) / Zn2+(aq) // H+(aq) / H2(g) / Pt(s)
Calculate the value of Ecell, in V, when [Zn2+] =1.00 M and [H+] = 0.00100 M
DATA:
Zn2+(aq) +2 e¯ ---> Zn(s);Eo = - 0.76 V
2 H+(aq) + 2 e¯ ---> H2(g);Eo = 0.00 V
The cell reaction is
Zn + 2H^+ ==> Zn^2+ + H2
Calculate Ecell = 0.76v + 0 = 0.76v.
Then Ecell = Eocell + (0.05916/2)*log(Q) whee Q = (pH2)*(Zn^2+)/(H^+)^2(Zn)(s)
I assume this is done at STP so pH2 = 1.00 atm and (Zn)(s) = 1.00 by definition. Plug in H^+ and Zn^2+ and solve for Ecell.
Well, it looks like we have a little chemistry problem here. Don't worry, I'm here to help you out with a side of humor.
In this voltaic cell, we have Zn(s) and H2(g) battling it out to see who can generate the most energy. It's like a miniature superhero showdown, but with electrons! Let's see who comes out on top.
First, let's find the oxidation half-reaction. Zn(s) is getting oxidized to Zn2+(aq), and it's not too happy about it with a standard reduction potential of -0.76 V. That value is like the Zn(s) saying, "Man, I wish I didn't have to give up these electrons."
On the other hand, H+(aq) is feeling pretty confident with a standard reduction potential of 0.00 V. It's like the H+(aq) saying, "I don't mind taking those electrons and turning into H2(g), it's a gas job!"
Now, let's find the reduction half-reaction. We flip the oxidation half-reaction for Zn(s), so it becomes Zn2+(aq) + 2 e¯ --> Zn(s).
Now, we need to balance the number of electrons in both half-reactions. We have 2 electrons in the reduction half-reaction and only 1 electron in the oxidation half-reaction. We can multiply the oxidation half-reaction by 2, like saying, "Hey Zn(s), we need more electrons over here!"
The final balanced equation is:
2 Zn(s) + 4 H+(aq) --> 2 Zn2+(aq) + 2 H2(g)
Now comes the part where we calculate the overall cell potential, Ecell. To do that, we need to sum up the standard reduction potentials of both half-reactions.
Ecell = E°reduction - E°oxidation
Ecell = 0.00 V - (-0.76 V)
Ecell = 0.76 V
So, the value of Ecell in this supercharged voltaic cell is 0.76 V when [Zn2+] = 1.00 M and [H+] = 0.00100 M. It's like the Zn(s) is saying, "Fine, H2(g), you win this round, but I'll be back with a better punchline next time!"
Hope that helps lighten up the chemistry conundrum a bit!
To calculate the value of Ecell for the given voltaic cell, we can use the formula:
Ecell = Eocathode - Eoanode
where Eocathode is the standard reduction potential of the cathode (H2/H+) and Eoanode is the standard reduction potential of the anode (Zn2+/Zn).
1. Determine the Eocathode and Eoanode using the given standard reduction potentials:
- Eocathode = 0.00 V
- Eoanode = -0.76 V
2. Calculate Ecell:
Ecell = Eocathode - Eoanode
= 0.00 V - (-0.76 V)
= 0.76 V
Therefore, the value of Ecell (cell potential) for the given voltaic cell with [Zn2+] = 1.00 M and [H+] = 0.00100 M is 0.76 V.
To calculate the value of Ecell for the given voltaic cell, we need to find the sum of the standard reduction potential for the reduction half-reaction and the standard reduction potential for the oxidation half-reaction.
The reduction half-reaction is:
Zn2+(aq) + 2 e¯ ---> Zn(s); Eo = -0.76 V
And the oxidation half-reaction is:
2 H+(aq) + 2 e¯ ---> H2(g); Eo = 0.00 V
Since the reduction half-reaction is multiplied by 2 for balancing the electrons transferred, we need to multiply its standard reduction potential by 2 as well.
Now, we can calculate the value of Ecell using the Nernst equation:
Ecell = Eocell - (0.0592 V / n) * log(Q)
Where:
- Eocell is the standard cell potential
- n is the number of electrons transferred in the balanced cell reaction
- Q is the reaction quotient, which is the ratio of the concentrations of the products and reactants raised to their stoichiometric coefficients
In this case, the standard cell potential (Eocell) is the sum of the standard reduction potentials for the two half-reactions:
Eocell = Eored + Eoox
Eocell = (-0.76 V) + (2 * 0.00 V)
Eocell = -0.76 V
Now, let's calculate the value of Ecell using the Nernst equation:
Ecell = (-0.76 V) - (0.0592 V / 2) * log[(Zn2+) / (H+)^2]
Given [Zn2+] = 1.00 M and [H+] = 0.00100 M, we can substitute these values into the equation:
Ecell = (-0.76 V) - (0.0592 V / 2) * log(1.00 M / (0.00100 M)^2)
Using logarithmic rules, we can simplify the equation:
Ecell = (-0.76 V) - (0.0592 V / 2) * log(1.00 M / 0.00000100 M^2)
Ecell = (-0.76 V) - (0.0592 V / 2) * log(1.00 / 0.00000100^2)
Ecell = (-0.76 V) - (0.0592 V / 2) * log(1.00 / 0.0000000001)
Ecell = (-0.76 V) - (0.0592 V / 2) * log(10^10)
Now, we can simplify further:
Ecell = (-0.76 V) - (0.0592 V / 2) * 10
Ecell = (-0.76 V) - (0.0592 V / 2) * 10
Ecell = (-0.76 V) - (0.0296 V) * 10
Ecell = -0.76 V - 0.296 V
Ecell = -1.056 V
Therefore, the value of Ecell for the given voltaic cell is -1.056 V.