Write the net ionic equations that occur in the following cells: Pb|Pb(NO_3)_2||
ID: 529528 • Letter: W
Question
Write the net ionic equations that occur in the following cells: Pb|Pb(NO_3)_2||AgNO_3|Ag Zn|ZnCl_2||Pb(NO_3)_2|Pb Pb|Pb(NO_3)_2||NiCl_2|Ni Which of the following reactions will have the larger emf under standard conditions? why? CuSO_4(aq) + Pb(s) PbSO_4(s) + Cu(s) Cu(NO_3)_2(aq) + Pb(s) Pb(NO_3)_2(aq) + Cu(s) Calculate Delta G for the reaction in Example 17.3. Voltages listed in textbooks and handbooks are given as a standard cell potentials (voltages). meant by a standard cell? Were the cells constructed in this experiment standard cell? Why or why As a standard voltaic cell runs, it "discharges" and the cell potential decreases with time. Explain. Using standard potentials given in the appendices, calculate the standard cell potentials and the for the following reactions:Explanation / Answer
2) CuSO4 (aq) + Pb (s) <=====> PbSO4 (s) + Cu (s)
Cu(NO3)2 (aq) + Pb (s) <=====> Pb(NO3)2 (aq) + Cu (s)
The emf of the cell is given as
Ecell = E0cell – 0.0591*log [Pb2+]/[Cu2+] …..(1)
PbSO4 is a sparingly soluble electrolyte and dissociates as below:
PbSO4 (s) <=====> Pb2+ (aq) + SO42- (aq)
Pb(NO3)2 is completely soluble in water and ionizes as below:
Pb(NO3)2 (aq) -------> Pb2+ (aq) + 2 NO3- (aq)
Since PbSO4 is sparingly soluble in water, [Pb2+] obtained by dissociation of PbSO4 will be low. On the contrary, Pb(NO3)2, being completely water soluble, will furnish a higher [Pb2+]. Greater the [Pb2+], higher (more positive or less negative) will be the value of the logarithmic term in expression (1) above. As the logarithmic term becomes more positive, the value of Ecell decreases. Therefore, PbSO4 will have a lower logarithmic term and hence a higher value of Ecell. Thus, the emf of the first reaction will be larger (ans).
3) Need the example 17.3 to answer the question.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.