This is the question I am having trouble with:
A voltaic cell consists of a strip of lead metal in a solution of Pb(NO3)2 in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with Cl2 gas bubbled around the electrode. A salt bridge connects the two beakers.
I have already figured out that the
--anode is: Pb
--cathode is: Pt
--and that the Pb electrode loses mass as the cell reaction proceeds.
I am having trouble writing "overall cell reaction"
(c) Write the equation for the overall cell reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
and
(d) What is the emf generated by the cell under standard conditions?
But I realize that I first need to get the correct reaction before I can do part D. If you can help me out that would be much appreciated! Thank You!!
If you know the anode and cathode already you can get the reaction from that. At the anode you have
Pb ==> Pb^+ + 2e
At the cathode you have
Cl2 + 2e ==> 2Cl^-
Add the two half equations for the cell reaction. You can add the phases.
Hi Dr. Bob222
Thank you so much for replying! I actually have already tried this equation...
Pb(s) + Cl2(g) --> Pb^2+(aq) 2Cl^-(aq)
and it said it was wrong
So that is why I am confused. Because I was initially solving it the way you had written the half reactions and then joining it together.
I don't know. I don't think it can be PbCl2 forming since the two beakers are connected only with a salt bridge. The cathode surely must be the Cl2 since the Pt is an inert electrode and no conditions are given for the H2O there (that is I don' think we would worry about the H2O reacting). That is the 2H2O + O2 + 4e --> 4OH^-
If you submitted the answer exactly as you posted it to me there is a + sign missing in the products and that could acount for the database saying it was wrong; however, I'm sure that's just a typo.
If that error message persists, I don't know where they are going. Sorry about that.
It turns out I had entered it in wrong! Thank you so much for your help! I really appreciate it!
whew! I'm glad I didn't goof.
You typed the anwser in mixed around. It should be Cl2+Pb--->[Pb}2+(aq)+2[Cl]-
Sure! Let's break down the process and figure out the overall cell reaction.
In a voltaic cell, oxidation occurs at the anode and reduction occurs at the cathode.
The anode in this case is the strip of lead metal (Pb), which is oxidized. The oxidation half-reaction can be written as follows:
Pb(s) ⟶ Pb²⁺(aq) + 2e⁻
On the other hand, the cathode is the platinum electrode (Pt), where reduction occurs. The reduction half-reaction can be written as follows:
2Cl⁻(aq) ⟶ Cl₂(g) + 2e⁻
Now, to write the overall cell reaction, we need to balance the number of electrons transferred in each half-reaction. To do this, we need to multiply the oxidation half-reaction by 2 so that both half-reactions involve the transfer of 2 electrons:
2Pb(s) ⟶ 2Pb²⁺(aq) + 4e⁻
2Cl⁻(aq) ⟶ Cl₂(g) + 2e⁻
Now, we can combine the two half-reactions to form the overall cell reaction:
2Pb(s) + 2Cl⁻(aq) ⟶ 2Pb²⁺(aq) + Cl₂(g)
This is the overall cell reaction for the given voltaic cell.
Now, let's move on to part D and determine the emf (electromotive force) generated by the cell under standard conditions.
To calculate the emf, we need to use the Nernst equation, which is given as follows:
emf = E°cell - (RT/nF) * ln(Q)
In this equation, E°cell represents the standard cell potential, R is the ideal gas constant, T is the temperature in Kelvin, n is the stoichiometric coefficient of electrons transferred in the overall cell reaction, F is the Faraday constant, and ln(Q) represents the natural logarithm of the reaction quotient.
To determine the standard cell potential (E°cell), you can look up the standard reduction potentials for the half-reactions involved in the Nernst equation. The standard reduction potential for the reduction of Cl⁻ to Cl₂ is available in reference tables.
To calculate the reaction quotient (Q), you need to consider the concentrations of the species involved in the reaction. Since you have the solutions and gases involved in each beaker, you can calculate their concentrations based on their molarity or partial pressure, respectively.
Once you have all the necessary values, you can plug them into the Nernst equation to calculate the emf generated by the cell under standard conditions.