A 60.0 mL solution of 0.00750M OSCI,3- in 1 M HCI was titrated with 0.0300 M IrCl,2- to give OsCl,2- and IrCl,3-. a) What is the reducing agent in this reaction? Write "a" for the titrant or "b" for the analyte. b) From the standard potentials table, what is the standard potential of the Os half-reaction? V c) What is the standard potential of the Ir half-reaction? E° V

Chemistry: Principles and Practice
3rd Edition
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Chapter18: Electrochemistry
Section: Chapter Questions
Problem 18.103QE
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Osmium
OsO,(8) + 8H* + 8e¯=Os(s) + 4H,O
OsCl + e= OsCl
0.834
-0.458
0.85
1F HCI
Iridium
IrCl
IrBr. + e = IrBr
IrCl + 4e = Ir(s) + 6CI
IrO,(s) + 4H* + 4e= Ir(s) + 2H,O
Irl, + e= Irl
+ e"= IrCl
1.026 1 F HCI
0.947 2 F NaBr
0.835
0.73
0.485 1 F KI
-0.36
Silver
2.000 4 F HCIO4
1.989
Ag** + e= Ag*
0.99
1.929 4 F HNO,
Ag* + 2e= Ag*
AgO(s) + H* + e=}Ag,O(s) + H‚O
Ag* + e= Ag(s)
Ag,C204(s) + 2e¯=2Ag(s) + C,0;-
AgN3(s) + e¯= Ag(s) + N5
AgCl(s) + e¯= Ag(s) + Cl¯
1.9
1.40
0.799 3
-0.989
0.465
0.293
0.222
0.197 saturated KCI
AgBr(s) + e= Ag(s) + Br¯
Ag(S,O3) + e¯= Ag(s) + 2S,Ož-
AgI(s) + e=Ag(s) + I¯
Ag,S(s) + H* + 2e=2Ag(s) + SH
0.071
0.017
-0.152
-0.272
Transcribed Image Text:Osmium OsO,(8) + 8H* + 8e¯=Os(s) + 4H,O OsCl + e= OsCl 0.834 -0.458 0.85 1F HCI Iridium IrCl IrBr. + e = IrBr IrCl + 4e = Ir(s) + 6CI IrO,(s) + 4H* + 4e= Ir(s) + 2H,O Irl, + e= Irl + e"= IrCl 1.026 1 F HCI 0.947 2 F NaBr 0.835 0.73 0.485 1 F KI -0.36 Silver 2.000 4 F HCIO4 1.989 Ag** + e= Ag* 0.99 1.929 4 F HNO, Ag* + 2e= Ag* AgO(s) + H* + e=}Ag,O(s) + H‚O Ag* + e= Ag(s) Ag,C204(s) + 2e¯=2Ag(s) + C,0;- AgN3(s) + e¯= Ag(s) + N5 AgCl(s) + e¯= Ag(s) + Cl¯ 1.9 1.40 0.799 3 -0.989 0.465 0.293 0.222 0.197 saturated KCI AgBr(s) + e= Ag(s) + Br¯ Ag(S,O3) + e¯= Ag(s) + 2S,Ož- AgI(s) + e=Ag(s) + I¯ Ag,S(s) + H* + 2e=2Ag(s) + SH 0.071 0.017 -0.152 -0.272
A 60.0 mL solution of 0.00750M OSCI63- in 1 M HCI was titrated with 0.0300 M IrCl,2- to give OsCl,2- and IrCl,3-.
a) What is the reducing agent in this reaction? Write "a" for the titrant or "b" for the analyte.
b) From the standard potentials table, what is the standard potential of the Os half-reaction?
E°+ =
c) What is the standard potential of the Ir half-reaction?
E°+ =
V
d) At what volume of analyte is the equivalence point?
Ve =
mL
e) Calculate the potential (vs. Ag | AgCl) when 10.8 mL of titrant is added.
E =
V
f) Calculate the potential (vs. Ag | AgCI) at the equivalence point.
E =
V
g) Calculate the potential (vs. Ag | AgCI) when 17.6 mL of titrant is added.
E =
h) At what volume of analyte will the potential be equal to the standard potential of the Ir half-reaction (ignoring the Ag | AgCl electrode)?
V =
mL
Transcribed Image Text:A 60.0 mL solution of 0.00750M OSCI63- in 1 M HCI was titrated with 0.0300 M IrCl,2- to give OsCl,2- and IrCl,3-. a) What is the reducing agent in this reaction? Write "a" for the titrant or "b" for the analyte. b) From the standard potentials table, what is the standard potential of the Os half-reaction? E°+ = c) What is the standard potential of the Ir half-reaction? E°+ = V d) At what volume of analyte is the equivalence point? Ve = mL e) Calculate the potential (vs. Ag | AgCl) when 10.8 mL of titrant is added. E = V f) Calculate the potential (vs. Ag | AgCI) at the equivalence point. E = V g) Calculate the potential (vs. Ag | AgCI) when 17.6 mL of titrant is added. E = h) At what volume of analyte will the potential be equal to the standard potential of the Ir half-reaction (ignoring the Ag | AgCl electrode)? V = mL
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