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Consider the following half-reactions:

IrCl63-+3e-Ir+6Cl-Ep=0.77VPrCl42-+2e-Pt+4Cl-Eo=0.73VPdCl42-+2e-Pd+4Cl-Eo=0.62V

A hydrochloric acid solution contains platinum, palladium, and iridium as chloro-complex ions. The solution is a constant 1.0 M in chloride ion and 0.020 M in each complex ion. Is it feasible to separate the three metalsfrom this solution by electrolysis? (Assume that 99% of metal must be plated out before another metal begins to plate out.)

Short Answer

Expert verified

It is possible to separate these metals by electrolysis. The reduction potential in decreasing order is Ir>Pt>Pd.

Step by step solution

01

Possible reduction reaction for metals

The reduction reaction for iridium can be denoted as:

IrCl63-+3e-Ir+6Cl-

The reduction potential is 0.77 V.

The reduction reaction for platinum is expressed as:

PtCl42-+2e-Pt+4Cl-

The reduction potential is 0.73 V.

The reduction potential for palladium can be denoted as:

PdCl42-+2e-Pd+4Cl-

The reduction potential is 0.62 V.

02

Separating these metals from the solution

There is a possibility of separating these metals from the solution when the voltage increases due to the non-overlapping of potential value. The reduction potential in decreasing order is Ir>Pt>Pd.

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Most popular questions from this chapter

Consider a concentration cell that has both electrodes made of some metal M. Solution A in one compartment of the cell contains 1.0 M M2+ Solution B in the other cell compartment has a volume of 1.00 L. At the beginning of the experiment, 0.0100 mole of M(NO3)2 and 0.0100 mole of Na2SO4, are dissolved in solution B (ignore volume changes), where the reaction
M2+(aq)+SO42-(aq)MSO4(s)
occurs. For this reaction equilibrium is rapidly established, whereupon the cell potential is found to be +0.44 V at 25°C. Assume that the process
M2++e-M
has a standard reduction potential of +0.80 V and that no other redox process occurs in the cell. Calculate the value of Ksp for MSO4(s) at 25°C.

You have a concentration cell in which the cathode hasa silver electrode with 0.10 M Ag+. The anode also hasa silver electrode with Ag+ (aq), 0.050 M, and 1.0 X 10-3 MAg(S2O3)23-. You read the voltage to be 0.76 V.

a. Calculate the concentration of Ag+ at the anode.
b. Determine the value of the equilibrium constant for the formation ofAg(S2O3)23-.

Ag+(aq)+2S2O32-(aq)Ag(S2O3)23-(aq)K=?

What is electrochemistry? What are redox reactions? Explain the difference between a galvanic cell and an electrolytic cell.

Look up the reduction potential for Fe3+ to Fe2+. Lookup the reduction potential for Fe2+ to Fe. Finally, look upthe reduction potential for Fe3+ to Fe. You should noticethat adding the reduction potentials for the first twodoes not give you the potential for the third. Why not?Show how you can use the first two potentials to calculate the third potential.

The saturated calomel electrode, abbreviated SCE, is often used as a reference electrode in making electrochemical measurements. The SCE is composed solution of calomel(Hg2Cl2). The electrolyte solution is saturated KCIESCEis +0.242V relative to the standard hydrogen electrode. Calculate the potential for each of the following galvanic cells containing a saturated calomel electrode and the given half-cell components as standard conditions. In each case indicate whether the SCE is the cathode or the anode. Standard reduction potentials are found in Table 11.1.

a.Cu2++2e-Cub.Fe3++e-Fe2+c.AgCI+e-Ag+CI-d.AI3+3e-AIe.Ni2+2e-Ni

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