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
E= ____ V?
The cell is not spontaneous or spontaneous.
Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentrations...
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...
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) +...
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 potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2 (0.12 M |Fe2 (0.0012 M) Fe(s) E2 =-0.440 V Efe/Fe = 0.339 V Cu2t/Cu Is the electrochemical cell spontaneous or not spontaneous Ecell V as written at 25 °C? not spontaneous spontaneous Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C. Pt(s) Sn2(0.0060 M), Sn4+(0.14 M) Fe3+(0.13 M), Fe2+(0.0056 M) Pt(s)...
Calculate the cell potential for the reaction below when the concentrations of ions are: [Ag+ ] = 0.010 M and [Cu2+] = 0.750 M, at 25 °C. Cu(s) + 2Ag+ (aq) --> Cu2+(aq) + 2Ag(s) Given the standard reduction potentials: Cu2+(aq) + 2e– → Cu(s) Eϴ = 0.34 V Ag+ (aq) + e– → Ag(s) Eϴ = 0.80 V (A) 0.35 V (B) 0.44 V (C) 0.46 V (D) 0.48 V (E) 0.57 V
Calculate the cell potential for the reaction as written at 25.00 °C, given that [Cr2+]=0.891 M and [Fe2+]=0.0120 M. Use the standard reduction potentials in this table. Cr(s)+Fe2+(aq)↽−−⇀ Cr2+(aq)+Fe(s) E= V
Pt(s) + Fe2+ (aq) - P+2+ (aq) + Fe(s) [Fe2+] = 0.0083 M [P+2+] = 0.030 M The cell is not spontaneous. O spontaneous. Cu(s) + 2 Ag+ (aq) Cu2+ (aq) + 2 Ag(s) (Cu²+] = 0.013 M [Ag+] = 0.013 M The cell is O not spontaneous O spontaneous. Co2+ (aq) + T18+ (aq) [Co2+] = 0.060 M (T3+1 = 0.0060 M Co + (aq) + Ti”+ (aq) Co+] = 0.050 M [Ti?+] = 0.0115 M E-L v The...
Calculate the cell potential for the following reaction as written at 25.00 °C, given that [Mg2 ] = 0.774 M and [Sn2 ] = 0.0190 M. Standard reduction potentials can be found here. Reduction Half-Reaction Standard Potential Ered° (V) F2(g) + 2e– → 2F–(aq) +2.87 O3(g) + 2H3O+(aq) + 2e– → O2(g) + 3H2O(l) +2.076 Co3+(aq) + e– → Co2+(aq) +1.92 H2O2(aq) + 2H3O+(aq) + 2e– → 2H2O(l) +1.776 N2O(g) + 2H3O+(aq) + 2e– → N2(g) + 3H2O(l) +1.766 Ce4+(aq) + e– → Ce3+(aq)...
26. Determine the cell potential under the stated conditions for the electrochemical reaction described. State whether each is spontaneous or nonspontaneous under the set of conditions at 298.15 K. Hg(l) + s (aq, 0.10 M) + 2Ag (aq, 0.25 M) - 2Ag(s) + HgS(s) E°= -0.70 V E°= 0.7996 V HgS(s) +2e → Hg(1) +S?(aq) Agt (aq) + e +Ag(s) A.Ecell = 1.43 V, spontaneous B. Ecel = 1.50 V, spontaneous C. Ecell = 0.0996 V, spontaneous on TT 1...
Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 °C Cu(s) Cu2+ (0.15 M) Fe2+ (0.0039 M) Fe(s) E =-0.440 V E+Cu = 0.339 V Fe2+/Fe Is the electrochemical cell spontaneous or not spontaneous -0.779 Ecell = as written at 25 C? not spontaneous spontaneous о Calculate the potential of the electrochemical cell and determine if it is spontaneous as written at 25 'C Pt(s) Sn2 (0.0024 M), Sn4+ (0.12 M) |...