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

1) Calculate ΔG∞ for the electrochemical cell Pb(s) | Pb2+(aq) || Fe3+(aq) | Fe2+(aq) | Pt(s)....

1) Calculate ΔG∞ for the electrochemical cell Pb(s) | Pb2+(aq) || Fe3+(aq) | Fe2+(aq) | Pt(s). A. –1.2 x 102 kJ/mol B. –1.7 x 102 kJ/mol C. 1.7 x 102 kJ/mol D. –8.7 x 101 kJ/mol E. –3.2 x 105 kJ/mol 2)Determine the equilibrium constant (Keq) at 25∞C for the reaction Cl2(g) + 2Br– (aq) 2Cl– (aq) + Br2(l). A. 1.5 x 10–10 B. 6.3 x 109 C. 1.3 x 1041 D. 8.1 x 104 E. 9.8

For the following electrochemical cell Cu(s)|Cu^2+ (aq, 0.0155 M)||Ag^+ (aq, 3.50 M)|Ag(s) write the net cell...

For the following electrochemical cell Cu(s)|Cu^2+ (aq, 0.0155 M)||Ag^+ (aq, 3.50 M)|Ag(s) write the net cell equation. Phases are optional. Do not include the concentrations. Calculate the following values at 25.0 °C using standard potentials as needed. Eocell delta Go reaction Ecell delta G reaction

Derive a balanced equation for the reaction occurring in the cell: Fe(s)|Fe2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) a.) If E?cell =...

Derive a balanced equation for the reaction occurring in the cell: Fe(s)|Fe2+(aq)||Fe3+(aq),Fe2+(aq)|Pt(s) a.) If E?cell = 1.21 V, calculate ?G? for the reaction. b.) If E?cell=1.21V, calculate the equilibrium constant for the reaction. c.) Use the Nernst equation to determine the potential for the cell: Fe(s)|Fe2+(aq,1.0�10-3M)||Fe3+(aq,1.0�10-3M),Fe2+(aq,0.10M)|Pt(s)

Consider an electrochemical cell based on the following overall reaction, Fe(s) + 2Ag+(aq)  Fe2+(aq) +...

Consider an electrochemical cell based on the following overall reaction, Fe(s) + 2Ag+(aq)  Fe2+(aq) + 2Ag(s) Fe2+(aq) + 2e– Fe(s), ℰ° = –0.44 V Ag+(aq) + e– Ag, ℰ° = 0.80 V Calculate the cell potential (in V) for this reaction at 25oC when the concentration of Ag+ ions is 0.050 M and the concentration of Fe2+ ions is 1.50 M. a. +1.32 V b. +1.50 V c. +1.20 V d. -1.32 V e. -1.20 V

Insoluble PbBr2(s) precipitates when solutions of Pb(NO3)2(aq) and NaBr(aq) are mixed. Pb(NO3)2(aq) + 2 NaBr(aq) →...

Insoluble PbBr2(s) precipitates when solutions of Pb(NO3)2(aq) and NaBr(aq) are mixed. Pb(NO3)2(aq) + 2 NaBr(aq) → PbBr2(s) + 2 NaNO3(aq) ΔrH° = ? To measure the enthalpy change, 200. mL of 0.75 M Pb(NO3)2(aq) and 200. mL of 1.5 M NaBr(aq) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises by 2.44 °C. Calculate the enthalpy change for the precipitation of PbBr2(s), in kJ/mol. (Assume the density of the solution is 1.0 g/mL, and its specific heat...

the equation: 2agn03 (aq) +cu (s) heat 2 ag (s) + Cu (NO3) 2 (aq) is...

the equation: 2agn03 (aq) +cu (s) heat 2 ag (s) + Cu (NO3) 2 (aq) is said to be written as?

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0...

Reaction is: Pb(NO3)2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO3(aq) 2.0 ml of 0.250M Pb(NO3)2 and 7.0 ml of 0.250M KI A) In your experiment, one of the reagents was limiting and the other was in excess. Therefore after the lead(II)iodide precipitated from the solution, there were spectator ions left in the solution, and also the unreacted excess reagent. Assuming complete precipitation of lead (II) iodide, calculate the concentrations of each type of ion (Pb2+, K+, NO3-, I-) present in the...

2. Consider the equilibria shown below: Pb3(PO4)2(s)→← 3 Pb2+(aq) + 2 PO43-(aq) &nbsp

2. Consider the equilibria shown below: Pb3(PO4)2(s)→← 3 Pb2+(aq) + 2 PO43-(aq) Keq = 1.0 x 10-54 Pb2+(aq) + 3 OH-(aq)→← Pb(OH)3-(aq) Keq = 8.0 x 1013 Pb3(PO4)2(s) + 9 OH-(aq)→← 3 Pb(OH)3-(aq) + 2 PO43-(aq) a) Calculate the equilibrium constant for the third equilibrium shown. b) Calculate the equilibrium concentrations of Pb(OH)3- and of PO43- in a pH 10 solution containing the third equilibrium.

An electrochemical cell, composed of an iron electrode (anode) immersed in 0.10 M Fe (NO3)2 and...

An electrochemical cell, composed of an iron electrode (anode) immersed in 0.10 M Fe (NO3)2 and silver electrode (cathode) immersed in 0.11M AgNO3, was found to have a cell potential of 1.21V. The standard reduction potential for Ag+ / Ag half-cell is +0.80V. The standard reduction potential for the Fe+2/ Fe half – cell is -0.45V. 1. Write the half-cell reaction at the anode. 2.Write the half –cell reaction at the Cathode 3.Write over all cell reaction 4. Determine the...

Consider the reaction: Li2S(aq) + Co(NO3)2 (aq) à 2 LiNO3 (aq) + CoS (s) What is...

Consider the reaction: Li2S(aq) + Co(NO3)2 (aq) à 2 LiNO3 (aq) + CoS (s) What is the mole-mole factor (written as a conversion factor) for each reactant to product CoS (s)? Calculate the molar mass of each substance in the above chemical reaction. If 4.2 g of Li2S are placed in a flask with excess Co(NO3)2, how many grams of CoS are expected? (This is theoretical yield) The reagents in this reaction are aqueous solutions. If 20 mL of 0.25...