Question
A student is given a standard galvanic cell, represented above, that has a Cu electrode and a Sn electrode. As current flows through the cell, the student determines that the Cu electrode increases in mass and the Sn electrode decreases in mass.
(a) Identify the electrode at which oxidation is occurring. Explain your reasoning based on the student’s observations.
(b) As the mass of the Sn electrode decreases, where does the mass go?
(c) In the expanded view of the center portion of the salt bridge shown in the diagram below, draw and label a particle view of what occurs in the salt bridge as the cell begins to operate. Omit solvent molecules and use arrows to show the movement of particles.
d) A nonstandard cell is made by replacing the 1.0 M solutions of Cu\((NO_3)_2\) and Sn\((NO_3)_2\) in the standard cell with 0.50 M solutions of Cu(NO3)2 and Sn\((NO_{3})_{2}\). The volumes of solutions in the nonstandard cell are identical to those in the standard cell.
(i) Is the cell potential of the nonstandard cell greater than, less than, or equal to the cell potential of the standard cell? Justify your answer.
(ii) Both the standard and nonstandard cells can be used to power an electronic device. Would the nonstandard cell power the device for the same time, a longer time, or a shorter time as compared with the standard cell? Justify your answer.
(e) In another experiment, the student places a new Sn electrode into a fresh solution of 1.0 M\( Cu(NO_{3})_{2}\) ·
(i) Using information from the table above, write a net-ionic equation for the reaction between the Sn electrode and the Cu\((NO_3)_2\) solution that would be thermodynamically favorable. Justify that the reaction is thermodynamically favorable.
(ii) Calculate the value of \(\Delta G\) for the reaction. Include units with your answer.
▶️Answer/Explanation
(a) Since the Sn electrode is losing mass, Sn atoms must be forming \(Sn^{2+}\)(aq). This process is oxidation. OR because the cell operates, E° must be positive and, based on the E° values from the table, it must be Sn that is oxidized.
(b) The atoms on the Sn electrode are going into the solution as \(Sn^{2+}\) ions.
(c) The response should show at least one \(K^+\) ion moving toward the Cu compartment on the left and at least one \(NO_{3}^-\) ion moving in the opposite direction.
d(i) It is the same. In the cell reaction \(Q= \frac{[Sn^{2+}]}{ [Cu^{2+}]}\) , and the concentrations of \(Sn^{2+}\) and \(Cu^{2+}\) are equal to each other in both cases.
d(ii) The nonstandard cell would power the device for a shorter time because the supply of \(Cu^{2+}\) ions will be exhausted more quickly. OR The nonstandard cell would power the device for a shorter time because the reaction will reach E=0 more quickly.
e (i) \(Cu^{2+}(aq) + Sn(s) → Cu(s) + Sn^{2+}\)(aq) E° is positive \((0.34 V + 0.14 V = 0.48 V)\), therefore the reaction is thermodynamically favorable. OR The cell observations from earlier parts of the question are evidence that the Sn is oxidized and Cu is reduced, therefore E° must be positive.
e.(ii) \(\Delta G^0=-nFE^0\)
\(\Delta G^0=-\frac{2 mol e^-}{mol_{rxn}}\times \frac{96,485}{mol e^-}\times \frac{0.48}{c}=-93,000 J/mol_{rxn}\)
Question
A student is given a standard galvanic cell, represented above, that has a Cu electrode and a Sn electrode. As current flows through the cell, the student determines that the Cu electrode increases in mass and the Sn electrode decreases in mass.
(a) Identify the electrode at which oxidation is occurring. Explain your reasoning based on the student’s observations.
(b) As the mass of the Sn electrode decreases, where does the mass go?
(c) In the expanded view of the center portion of the salt bridge shown in the diagram below, draw and label a particle view of what occurs in the salt bridge as the cell begins to operate. Omit solvent molecules and use arrows to show the movement of particles.
(d) A nonstandard cell is made by replacing the 1.0 M solutions of Cu(NO3)2 and Sn(NO3)2 in the standard cell with 0.50 M solutions of Cu(NO3)2 and Sn(NO3)2 . The volumes of solutions in the nonstandard cell are identical to those in the standard cell.
(i) Is the cell potential of the nonstandard cell greater than, less than, or equal to the cell potential of the standard cell? Justify your answer.
(ii) Both the standard and nonstandard cells can be used to power an electronic device. Would the nonstandard cell power the device for the same time, a longer time, or a shorter time as compared with the standard cell? Justify your answer.
(e) In another experiment, the student places a new Sn electrode into a fresh solution of 1.0 M Cu(NO3)2 .
(i) Using information from the table above, write a net-ionic equation for the reaction between the Sn electrode and the Cu(NO3)2 solution that would be thermodynamically favorable. Justify that the reaction is thermodynamically favorable.
(ii) Calculate the value of \(\Delta G°\) for the reaction. Include units with your answer.
▶️Answer/Explanation
Ans:
(a)
Since the Sn electrode is losing mass, Sn atoms must be forming Sn2+(aq). This process is oxidation. OR because the cell operates, E0 must be positive and, based on the E0 values from the table, it must be Sn that is oxidized. |
(b)
The atoms on the Sn electrode are going into the solution as Sn2+ ions |
(c)
The response should show at least one K+ ion moving toward the Cu compartment on the left and at least one NO3– ion moving in the opposite direction. |
(d) (i)
It is the same. In the cell reaction \(Q = \frac{[Sn^{2+}]}{[Cu^{2+}]},\) and the concentrations of Sn2+ and Cu2+ are equal to each other in both cases. |
(ii)
The nonstandard cell would power the device for a shorter time because the supply of Cu2+ ions will be exhausted more quickly. OR The nonstandard cell would power the device for a shorter time because the reaction will reach E = 0 more quickly. |
(e) (i)
Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq) E0 is positive (0.34 V + 0.14 V = 0.48 V), therefore the reaction is thermodynamically favorable. OR The cell observations from earlier parts of the question are evidence that the Sn is oxidized and Cu is reduced, therefore E0 must be positive. |
(ii)
DG0 = -nFE0 \(\Delta D^{0}= – \frac{2 mol e^{-}}{mol_{rxn}}\times \frac{96,485 C}{mol e^{-}}\times \frac{0.48 J}{C}=-93,000 J/mol_{rxn}=-93 kJ/mol_{rxn}\) |