An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3...

An ideal Pb2+Pb2+ ion-selective electrode is moved from a 5.69×10−55.69×10−5 M Pb(NO3)2Pb(NO3)2 solution to a 3.22×10−33.22×10−3 M Pb(NO3)2Pb(NO3)2 solution at 25 °C. By how many millivolts will the potential of the ion-selective electrode change when it is moved from the first solution to the second solution?

Solutions

Expert Solution

The potential measured by the electrode is given by:

Where R and F are constants (the universal and Faraday's constants, respectively), T is the temperature (298 K in this case), n is the number of electrons involved in the redox process (2 in this due, being this a divalent cation) and E° is the standard potential of the cell. If we consider all of these constants, the temperature, and the change from ln to log, we get:

So the change in the measured potential between the two concentrations will be:

queen_honey_blossom answered 1 year ago

Next > < Previous

Related Solutions

A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode...

A galvanic cell consists of a lead electrode in 0.0256 M Pb(NO3)2 and a aluminum electrode in 0.292 M Al(NO3)3. What is the cell potential (emf, in V) of this cell at 25oC?

If we mix 19.0 mL of a 1.30 ✕ 10−3 M solution of Pb(NO3)2 with 18.6...

If we mix 19.0 mL of a 1.30 ✕ 10−3 M solution of Pb(NO3)2 with 18.6 mL of a 2.60 ✕ 10−3 M solution of NaBr, will a precipitate form? At this point we would identify the potential precipitate by switching the anions and using the solubility rules to decide if one of the potential products is potentially insoluble. Here the products would be PbBr2 and NaNO3. Of these, PbBr2 is considered "insoluble". The Ksp value for PbBr2 is 6.60 ...

In the laboratory, an aqueous solution that is 4.00 M in Pb(NO3)2 is slowly added from...

In the laboratory, an aqueous solution that is 4.00 M in Pb(NO3)2 is slowly added from a buret to an aqueous solution that is 0.005 M in Cl- and also 0.0500 M in Br-. PbCl2(s) <-> Pb+2(aq) + 2 Cl-(aq) , Ksp = 1.6 x 10-5 PbBr2 (s) <-> Pb+2(aq) + 2 Br-(aq) , Ksp = 4.0 x 10-5 1. Which ion, Cl- or Br-, is the first to precipitate from solution? 2. When the second ion begins to precipitate,...

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added...

A solution is 0.010 M in each of Pb(NO3)2, Mn(NO3)2, and Zn(NO3)2. Solid NaOH is added until the pH of the solution is 8.50. Which of the following is true? Pb(OH)2, Ksp= 1.4 x10^-20 Mn(OH)2, Ksp= 2.0 x 10^-13 Zn(OH)2, Ksp= 2.1 x10^-16 A) No precipitate will form. B) Only Pb(OH)2 will precipitate. C) Only Mn(OH)2 wil precipitate. D) Only Zn(OH)2 and Pb(OH)2 will preipitate. E) All three hydroxide will preipitate.

What is the concentration (in molality) of lead nitrate (Pb(NO3)2) in a 0.798 M solution? The...

What is the concentration (in molality) of lead nitrate (Pb(NO3)2) in a 0.798 M solution? The density of the solution is 1.27 g/mL. Answer has 3 decimal places.

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of...

A solution is prepared by mixing 50 mL of 1 M Pb(NO3)2 and 75 mL of .5 M NaF. Calculate the concentration of F- ions present at equilibrium. (Hint: First, write and balance the double displacement reaction taking place between Pb(NO3)2 and NaF to form PbF2(s). Perform the necessary stoichiometry (including finding which reactant is limiting) to calculate how much of each reactant remains after the reaction goes to completion. Remember that stoichiometry has to be done in moles. Once...

A solution is prepared by mixing 100 mL of .01 M Pb(NO3)2 and 100 mL of...

A solution is prepared by mixing 100 mL of .01 M Pb(NO3)2 and 100 mL of .001 M NaF. Will PbF2(s) precipitate in this reaction? Calculate the concentration of F- ions present at equilibrium. PbF2(s) <--> Pb2+(aq) + 2F-(aq) Kc= 3.7E-8 Please show all work! Thank you!

Calculate [H+] for a 5.69×10-2 M aqueous solution of chloroacetic acid (HC2H2ClO2; Ka = 1.4×10-3). Calculate...

Calculate [H+] for a 5.69×10-2 M aqueous solution of chloroacetic acid (HC2H2ClO2; Ka = 1.4×10-3). Calculate the pH of the above solution. Thanks!

A solution is made that is 1.1×10−3 M in Zn(NO3)2 and 0.140 M in NH3. Part...

A solution is made that is 1.1×10−3 M in Zn(NO3)2 and 0.140 M in NH3. Part A After the solution reaches equilibrium, what concentration of Zn2+(aq) remains? Express your answer using two significant figures.

Answer the following for the reaction: Pb(NO3)2(aq)+2KCl(aq)→PbCl2(s)+2KNO3(aq) How many milliliters of a 2.05 M Pb(NO3)2 solution...

Answer the following for the reaction: Pb(NO3)2(aq)+2KCl(aq)→PbCl2(s)+2KNO3(aq) How many milliliters of a 2.05 M Pb(NO3)2 solution will react with 40.5 mL of a 1.45 M KCl solution? Please show work! Thanks

Subjects