Please show all work! Thanks! Calculate the equilibrium constants of the following reactions at 25°C from...
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6. Calculate the equilibrium constant at 25°C for the following reaction: Cd(s) + Sn** (aq) → Cd²+ (aq) + Sn(s) K = 6.3 x 108
19.03 pts Question 10 Use the standard reduction potenital to calculate equilibrium constants for the following reaction at 298.15 K. AgCl (s) Ag+ (aq) +Cl- (aq) Ag+ (aq) +Ag(s) E° = 0.7996 V AgCl(s) + e - Ag(s) + CH- (aq) E° = 0.22233 V O K=5.2 x 105 spontaneous O K = 3.8 x 10"nonspontaneous K - 1.7 x 10-10 nonspontaneous
Calculate the equilibrium constant for each of the reactions at 25∘C. Standard Electrode Potentials at 25 ∘C Reduction Half-Reaction E∘(V) Fe3+(aq)+3e− →Fe(s) -0.036 Sn2+(aq)+2e− →Sn(s) -0.14 Ni2+(aq)+2e− →Ni(s) -0.23 O2(g)+2H2O(l)+4e− →4OH−(aq) 0.40 Br2(l)+2e− →2Br− 1.09 I2(s)+2e− →2I− 0.54 A) 2Fe3+(aq)+3Sn(s)→2Fe(s)+3Sn2+(aq) (answers are not 4.1x10^5, 3.3x10^3, 2.7x10^10, or 2.6x10^10) B) O2(g)+2H2O(l)+2Ni(s)→4OH−(aq)+2Ni2+(aq) C) Br2(l)+2I−(aq)→2Br−(aq)+I2(s) (answer is not 1.7x10^18)
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9.15 Consider a hydrogen electrode in HBr(aq) Estimate the change in the electro de potential when the solution is changed from 5.0 mmol dm-3 to at 25°C operating at 1.45 bar. 15.0 mmol dm3 9.16 Devise a cell in which the cell reaction is: Mn(s) + Cl2(g) MnCl2(aq). Give the half-reactions for the electrodes and from the standard cell potential of +2.54 V deduce the standard potential of the Mn2/Mn couple. 9.17 Write the cell reactions, electrode half-reactions...
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6. The following equilibrium constants have been determined for hydrosulfuric acid at 25°C: H2S(aq)<> H(aq) + HS (aq) K. = 9.5 x 10-8 HS (aq) → H+ (aq) + S2 (aq) K". = 1.0 x 10-19 Calculate the equilibrium constant for the following reaction at the same temperature: H2S(aq) → H(aq) + S2- (aq)
Question 10 3.03 pts Use the standard reduction potenital to calculate equilibrium constants for the following reaction at 298.15 K. AgCl(s) Ag+ (aq) +CH(aq) Ag+ (aq) +eAg(s) E° =0.7996 VAgCl(s) + e - Ag(s) + CH(aq) E° = 0.22233 V OK-5.2 x 10% spontaneous OK=3.8 x 10nonspontaneous N O K=1.7 x 10-10 nonspontaneous
Post Lab Questions: To receive full credit, you must SHOW ALL YOUR WORK!! Use the table of reduction half reactions to answer the following Post-Lab questions: Table 3. Example reduction Reduction Potential Chart half reactions. The easiest to Ce+(aq) + 3e - Ce3+ (aq) reduce is at the top. The more Au3+ (aq) + 3e Au(s) difficult to reduce is at the bottom. Cl2(g) + 2e 2CH(ag) Ag+ (aq) + e- Ag(s) Fe3+ (aq) + e- Fe2+ (aq) AgCl(s) +...
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11. An electrochemical cell uses the following balanced chemical equation 3Ag (aq)+Al(s)AP(aq) + 3Ag(s) Ecell = +2.46 V A. (6 pts) Clearly label the following cell diagram with the following information: a. Anode and cathode b. Direction of electron flow under standard state conditions Direction of cation flow in the salt bridge c. AI Ag (s) (s) APtaa Aglaq) NO,(aq) NO (aq) B. (7 pts) What is the...
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4. Consider an electrochemical cell (a.k.a. galvanic cell or voltaic cell) with Ag(s) and 1.0 M AgNO3(aq) in one compartment and Cu(s) and 1.0 M Cu(NO3)2(aq) in the other compartment. Write the reactions and calculate the standard state cell potential at 298 K. E cathode = a. Reduction (cathode): Eanode = b. Oxidation (anode): Eºcell = C. Net (overall cell reaction): 5. Consider a galvanic cell with Sn(s) and 1.0 M Sn(NO3)2(aq) in one...
A voltaic cell is based on the following two half-reactions: Cd2 (ag) +2e-> Cd (s) Sn2(aq)+ 2e Sn (s) Calculate the standard cell potential. Use the date from the attached table.SRP2.docx Oa 0.13 Ob 042 Oc.027 Od-0.27