Asked by sarah
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.)
Write the equation for the overall cell reaction. (Use the lowest possible coefficients. Include states-of-matter under SATP conditions in your answer.)
Answers
Answered by
DrBob222
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.
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.
Answered by
sarah
y don't you use Pt for the second equation?
Answered by
sarah
would it be this?
Cd(s) + Cl2(g) + 2e==> Cd^2+(aq) + 2e + 2Cl^-(aq)
Cd(s) + Cl2(g) + 2e==> Cd^2+(aq) + 2e + 2Cl^-(aq)
Answered by
DrBob222
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 = ?
The equation you wrote is correct if you omit the 2e on each side. The voltage is 1.360 + 0.403 = ?
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