[16) The following question should be answered with reference to the diagram at

Question

[16) The following question should be answered with reference to the diagram at the right and appropriate standard reduction potentials on the cover page. 6. (a) Which electrode is the anode? What is the value for 0° when [Ag-[Cu 7-1.0 M? salt bridge Ag Cu Cu tog (b) Circle the quantity given below that increases immediately if the surface area of the silver electrode is increased. overall cell voltage change in ratio of electrode masses, ?-(mass of Cu/mass of Ag) for this cell if [Cu*]-1.00 M and [Ag1-0.010 M mass of Cu electrode rate of change of [Ag ] (c) Determine the value of the second term in the Nernst equation, (0.05916 V/n) log Q. (d) Circle the change that would increase the potential the most for this cell. doubling the [Ag] halving the [Cu?] decreasing the size of the Ag electrode by one half the potential of the cell can't be changed doubling the size of the Cu(s) electrode (e) What is the minimum cell voltage that must be exceeded to reverse the reaction occurring in this lectrochemical cell?

Explanation / Answer

a) The electrode potentials of Cu and Ag can be shown as follows:

Cu2+ + 2e- -----> Cu, E = 0.34V

Ag+ + e- -----> Ag, E = 0.80 V

Since oxidation occurs at an anode, and copper has higher oxidation potential(lower reduction potential), It would be copper electrode at the anode.

We can find out DG° using the formula:

DG° = -nFE°

where n is no of electrons involved in the cell:

The half reactions would be:

Cu2+ + 2e- -----> Cu,

2Ag -----> 2Ag+ + 2e-,

So the number of electrons involved would be 2, leaving n=2.

and F = 96.5 KJ/mol e-.V

E°cell = 0.80 V - 0.34V = 0.46V

DG° = -2* 96.5 KJ/mol.e-.V* 0.46V = -88.8 KJ/mol

b) The correct option is the rate of change of [Ag+]. This is because, increasing the surface area of the silver electrode increases the amountof Ag and hence the production of Ag+ ions.

c) The nernst equation is E = E0 - (0.05916V/n) log Q

n is no of electrons, in this case, n = 2.

Q = [Cu2+]/[Ag+] = 1.00/0.01 = 100

- (0.05916V/n) log Q = - (0.05916V* /2)log100 = - (0.05916V* /2)2 = -0.05916V

d) The cell potential can be increased the most by doubling the [Ag+]. As doublling this concentration will decrease the value of Q and hence increases E.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.