A voltaic cell consists of a strip of cadmium metal in a solution of Cd(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.
Write the equation for the overall cell reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
would it be this?
Cd(s) + Cl2(g) + 2e==> Cd^2+(aq) + 2e + 2Cl^-(aq)
At the anode (Cd) you have
Cd ==> Cd^2+ + 2e with Eo = 0.403 (as written) which is an oxidation.
At the cathode (Pt/Cl2) you have
Cl2 + 2e ==> 2Cl^- with Eo = 1.36 (as written) which is a reduction.
Add the two equation to obtain the cell reaction and add the two Eo values to obtain the cell potential. You may add the phases to the reaction.
y don't you use Pt for the second equation?
Pt is an inert electrode and is used because Cl2, being a gas, is not conductive.
The equation you wrote is correct if you omit the 2e on each side. The voltage is 1.360 + 0.403 = ?
To write the equation for the overall cell reaction, we need to consider the half-reactions happening at each electrode. Here's the step-by-step process to determine the overall cell reaction:
1. Identify the oxidation half-reaction: In this voltaic cell, the strip of cadmium metal in the Cd(NO3)2 solution is oxidized. Since cadmium is losing electrons, it is undergoing oxidation. Writing the half-reaction, we have:
Cd(s) -> Cd2+(aq) + 2e-
2. Identify the reduction half-reaction: The platinum electrode in the NaCl solution is the site of reduction. In this case, Cl2 gas is being reduced to Cl- ions. Writing the half-reaction, we have:
Cl2(g) + 2e- -> 2Cl-(aq)
3. Balance the half-reactions: Balance the number of atoms on each side of the half-reactions, and balance the charges by adding appropriate electrons. In this case, since there are 2 electrons in the oxidation half-reaction, we need to multiply the reduction half-reaction by 2 to balance the electrons. The balanced half-reactions then become:
Cd(s) -> Cd2+(aq) + 2e-
2Cl2(g) + 4e- -> 4Cl-(aq)
4. Combine the half-reactions: To form the overall cell reaction, add together the balanced half-reactions. Ensure that the number of electrons on both sides cancels out. In this case, we need to multiply the oxidation half-reaction by 2 to balance the electrons. The overall cell reaction is:
2Cd(s) + 4Cl2(g) -> 2Cd2+(aq) + 4Cl-(aq)
Therefore, the equation for the overall cell reaction is:
2Cd(s) + 4Cl2(g) -> 2Cd2+(aq) + 4Cl-(aq)