Pt(s) + Fe2+ (aq) - P+2+ (aq) + Fe(s) [Fe2+] = 0.0083 M [P+2+] = 0.030...
Question 7 of 16 > Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E ). Pt(s) + Fe2+ (aq) = P22+ (aq) + Fe(s) [Fe2+] = 0.0017 M [P12+] = 0.040 M The cell is O spontaneous. O not spontaneous Cu(s) + 2 Ag+ (aq) = Cu2+(aq) + 2 Ag(s) [Cu2+] =...
Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E Pt(s)Fe2+(aq) Pt2+(aq) + Fe(s) [Fe2+ [Pt2+] = 0.0023 M = 0.011 M The cell is V E = O not spontaneous. O spontaneous Cu(s)2 Ag (aq) Cu2+(aq) + 2 Ag(s) [Cu2+0.031 M [Ag*] = 0.031 M The cell is V E = O...
26. Determine the redox reaction represented by the following cell notation. Fe(s) I Fe2+(aq) I| Cu2+(aq) I Cu(s) A) Cu(s) + Fe2+(aq) B) Fe(s)+Cu2+(aq) Cu(s)+ Fe2+(aq) C) 2 Fe(s)+ Cu2+(aq)Cu(s) + 2 Fe2+(aq) D) 2 Cu(s) + Fe2+(aq)Fe(s)+ 2 Cu2+(aq) E) 3 Fe(s) + 2 Cu2+(aq) 2 Cu(s)+3 Fe2+(aq) Fe(s) + Cu2+(aq)
Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. QUESTION 1) Pt(s)+Fe2+(aq)−⇀↽−Pt2+(aq)+Fe(s) [Fe2+]=0.0013 M[Pt2+]=0.048 M E= ____ V? The cell is not spontaneous or spontaneous. QUESTION 2) Cu(s)+2Ag+(aq)−⇀↽−Cu2+(aq)+2Ag(s) [Cu2+]=0.017 M [Ag+]=0.017 M E= ____ V? The cell is not spontaneous or spontaneous. QUESTION 3) Co2+(aq)+Ti3+(aq)−⇀↽−Co3+(aq)+Ti2+(aq) [Co2+] = 0.065 M [Co3+] = 0.025 M [Ti3+] = 0.0060 M [Ti2+] = 0.0118 M...
For the electrochemical cell, Fe(s)| Fe2+ (aq)|| Cu2+ (aq)| Cu+ (aq) | Pt(s), determine the equlibrium constant (Keq) at 25°C for the reaction that occurs.
Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations provided. Then, determine if the cells are spontaneous or nonspontaneous as written. Refer to the table of standard reduction potentials (E Pt(s) Fe2+(aq) Pt2+(aq) + Fe(s) Fe2+0.0050 M Pt2+10.035 M The cell is E = V not spontaneous O spontaneous. Cu(s)2 Ag(aq) Cu2t(aq) + 2 Ag(s) [Cu2+0.015 M [Ag 0.015 M The cell is E = V O spontaneous. O not spontaneous Co2+(aq)Ti3(aq)Co3t(aq) +...
Half-Reaction Fe2+(aq) + 2e Fe(s) Hg2+ (aq) + 2e Hg() Ag+ (aq) + e + Ag(s) Cu2+ (aq) + 2e + Cu(s) Zn2+ (aq) + 2e → Zn(s) E (V) -0.44 0.86 0.80 0.34 - 0.76 Using the table, calculate Eºcell for the following electrochemical cell under standard conditions voltmeter a) 1.24 V Fe. salt bridge Ag b) -1.24 V c) 2.04 V d) - 2.04 V Ag a b С
question iii
10. An electrochemical cell consists of a standard Fe/Fe electrode (Fe (aq) (1.OM) IFe (aq) (1.0M) IPt(s) and a copper metal electrode ( concentrations) Cu2 (aq) +2e Cu(s) Fe (aq) le Fe (aq) +0.34 V -034 +0.77V Mark what is the corect balanced equation for the spontaneous reac u (aq) Fe (aq) Cu(s)+Fe2 (aq) C) Cu(s) + Fe (aq) Cu (aq) + Fe (aq) α Cu2+(aq) + 2e → Cu(s) + Fe2+(aq) Cu → Cu2ト+20 94 i) (...
1- Consider the following redox reaction: Fe(s) + Cu2+(aq) --> Fe2+(aq) + Cu(s) EoCell = 0.78 V If [Cu2+] = 0.3 M, what [Fe2+] is needed so that Ecell = 0.76 V 2- Calculate the Eocell for the following redox reaction: Cu(s) + 2Ag+(aq) --> Cu2+(aq) + 2Ag(s)
7) Calculate the cell potential of the following cell at 25°C. (7 points) Fe(s) Fe2+(aq) (1.1 M) || Cu2+ (aq) (0.50 M) Cu(s) Ecell = Eºcell - 0.0592/n logQ