Pt(s) ------------> Pt^2+ (aq) +2e^- E0 = -1.20v
Fe^2+ (aq) + 2e^- ------> Fe(s) E0 = -0.41v
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Pt(s) + Fe^2+ (aq) -----> Pt^2+ (aq) + Fe(s) E0cell = -1.61v
E0cell = -1.61v
n = 2
E = E0cell - 0.0592/n logQ
= -1.61-0.0592/2log[Pt^2+]/[Fe^2+]
= -1.61-0.0296log(0.04/0.0017)
=-1.61-0.029*1.3716
= -1.65v
E0cell<0
It is non spontaneous
Cu(s) --------> Cu^2+ (aq) + 2e^- E0 = -0.34v
2Ag^+ (aq) + 2e^- ------> 2Ag(s) E0 = 0.80v
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Cu(s) + 2Ag^+ (aq) -----> Cu^2+ (aq) + 2Ag(s) E0cell = 0.46v
n = 2
E = E0cell - 0.0592/n logQ
= 0.46-0.0592/2log[Cu^2+]/[Ag^+]^2
= 0.46-0.0296log(0.021/(0.021)^2
= 0.46 -0.0296*1.678
= 0.41v
E0cell>0
It is spontaneous
Co^2+ (aq) ----------------> Co^3+ (aq) + e^- E0 = -1.82v
Ti^3+ (aq) + e^- -----------------> Ti^2+ (aq) E0 = -0.90v
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Co^2+ (aq) + Ti^3+ (aq) ------> Co^3+ (aq) + Ti^2+ (aq) E0cell = -2.72v
n = 1
E = E0cell - 0.0592/n logQ
= -2.72 - 0.0592/1log[Co^3+][Ti^2+]/[Co^2+][Ti^3+]
= -2.72-0.0592 log(0.04*0.0098/0.05*0.0045)
=-2.72 -0.0592*0.241
= -2.734v
E0cell<0
it is non spontaneous
Question 7 of 16 > Calculate the cell potential, E, for the given reactions at 25.00 °C using the ion concentr...
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) +...
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...
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...
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Based on the sign of the standard cell potential, Ecell, classify these reactions as spontaneous or nonspontaneous as written. Assume standard conditions. Refer to the list of standard reduction potentials. Spontaneous as written Nonspontaneous as written Answer Bank Au + (aq) + 3 Ag(s) Au(s) + 3 Ag* (aq) 1,(s) + Cu() — 21(aq) + Cu2+ (aq) Ni2+ (aq) + Pb(s) Ni(s) + Pb2+ (aq)