Question

In: Chemistry

the voltage for the electrochemical cell shown below was .450 V. Pt(s)|Cd(NO3)2 (aq)(.0100M) || Cl- (aq)...

the voltage for the electrochemical cell shown below was .450 V.

Pt(s)|Cd(NO₃)₂ (aq)(.0100M) || Cl^- (aq) | AgCl (s)| Ag(s)

The standard reduction potentials are

Cd²⁺+2e^---> Cd (s) E^o= -.402 V

AgCl+e^---> Ag(s) + Cl^-E^o=.222V

Calcuate the concentration of Cl^- if the cell potential was determined to be 0.605 V

Expert Solution

Cd(s) -------------------> Cd^2+ (aq) + 2e^- E0 = 0.402v

2AgCl(s) + 2e^- ----------> 2Ag(s) + 2Cl^- (aq) E0 = 0.222v

-----------------------------------------------------------------------------------

Cd(s) + 2AgCl(s) --------------> Cd^2+ (aq) + 2Ag(s) + 2cl^- E0cell = 0.45v

n = 2

ECell = E0cell - 0.0592/n log[Cd^2+]Cl^-]^2

0.605 = 0.45 - 0.0592/2 log0.01[Cl^-]^2

0.605-0.45 = -0.0296log0.01[Cl^-]^2

0.155 = -0.0296log0.01[Cl^-]^2

log0.01[Cl^-]^2 = -0.155/0.0296

log0.01[Cl^-]^2 = -5.24

0.01[Cl^-]^2 = 10^-5.24

0.01[Cl^-]^2 = 5.75*10^-6

[Cl^-]^2 = 5.75*10^-6/0.01

[Cl^-]^2 = 5.75*10^-4

[Cl^-] = 2.4*10^-2 M >>>>answer

queen_honey_blossom answered 2 years ago

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