Consider the cell described below at 297 K: Fe | Fe2+ (0.791 M) || Cd2+ (0.953...
Consider the cell described below at 297 K:
Fe | Fe2+ (0.791 M) || Cd2+ (0.953 M) | Cd
Given EoCd2+→Cd = -0.403 V, EoFe2+→Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.349 mol/L.
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Consider the cell described below at 297 K:
Fe | Fe2+ (0.791 M) || Cd2+ (0.953...
Consider the cell described below at 253 K: Fe | Fe2+ (0.867 M) || Cd2+ (0.945...
Consider the cell described below at 253 K: Fe | Fe2+ (0.867 M) || Cd2+ (0.945 M) | Cd Given EoCd2+Cd -0.403 V, E°Fe2+_Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.311 mol/L. Tries 0/45 Submit Answer
Consider the cell described below at 275 K: Fel Fe2+ (0.935 M) || Cd2+ (0.939 M)...
Consider the cell described below at 275 K: Fel Fe2+ (0.935 M) || Cd2+ (0.939 M) 1 Cd Given E°Cd2+--Cd = -0.403 V, EºFe2+-Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.399 mol/L.
Consider the cell described below at 253 K: Sn | Sn2+ (0.979 M) || Pb2+ (0.943 M) | Pb Given E°pb2+_Pb = -0.131 V, Eºsn...
Consider the cell described below at 253 K: Sn | Sn2+ (0.979 M) || Pb2+ (0.943 M) | Pb Given E°pb2+_Pb = -0.131 V, Eºsn2+_Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed 0.355 mol/L. .00893 V Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previous Tries
HHelp ASAP Consider the cell described below at 283 K: Sn | Sn2+ (0.871 M) || Pb2+ (0.927 M) | Pb Given E°Pb2+>Pb = -...
HHelp ASAP
Consider the cell described below at 283 K: Sn | Sn2+ (0.871 M) || Pb2+ (0.927 M) | Pb Given E°Pb2+>Pb = -0.131 V, E°şn2+-Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.355 mol/L. -2780 V Incorrect. Tries 6/45 Previous Tries Submit Answer
Using the Nernst equation calculate the cell voltage for: Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s)...
Using the Nernst equation calculate the cell voltage for: Fe(s) + Cd2+(aq) → Fe2+(aq) + Cd(s) when the [Fe2+] = 0.20 M and [Cd2+] = 1.5 M. Potentially useful information: Fe2+ + 2e− → Fe(s); ε0 = -0.44 V Cd2+ + 2e− → Cd(s); ε0 = -0.40 V
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) +...
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) e°(Fe2+ / Fe) = -0.4400 V, Eº (Cd2+ / Ca) = -0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+1 0.75 M and Ca2+] = 0.010 M. E- L v Will the reaction occur spontaneously at these conditions? O yes no cannot predict
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) +...
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E o (Fe2+ / Fe) = −0.4400 V, E o (Cd2+ / Cd) = −0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.60 M and [Cd2+] = 0.010 M. E =____ V Will the reaction occur spontaneously at these conditions? yes no cannot predict
Consider a galvanic cell based on the following two half reactions under standard conditions at 298...
Consider a galvanic cell based on the following two half reactions under standard conditions at 298 K: Fe2+ + 2 e → Fe -0.440 V Cd2+ + 2 e → Cd -0.403 V What will the potential of the cell be when the cathode solution concentration changes by 0.657 M? Cell potential =
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) +...
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E° (Fe2+ / Fe) =-0.4400 V, Eº (Cd2+ / Cd) = -0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.85 M and [Cd2+] = 0.010 M. E= Lv Will the reaction occur spontaneously at these conditions? o o yes no cannot predict o
For the galvanic (voltaic) cell Cd2+ (aq) + Fe(s) Cd(s) + Fe2+(aq) (E° = 0.0400 V),...
For the galvanic (voltaic) cell Cd2+ (aq) + Fe(s) Cd(s) + Fe2+(aq) (E° = 0.0400 V), what is the ratio [Fe2+1/[Cd2+] when E = 0.001 V? Assume T is 298 K 1 2. 3 Х 4 5 6 C 7 8 9 + +/- 0 x 100
Consider the cell described below at 253 K: Fe | Fe2+ (0.867 M) || Cd2+ (0.945 M) | Cd Given EoCd2+Cd -0.403 V, E°Fe2+_Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.311 mol/L. Tries 0/45 Submit Answer
Consider the cell described below at 275 K: Fel Fe2+ (0.935 M) || Cd2+ (0.939 M) 1 Cd Given E°Cd2+--Cd = -0.403 V, EºFe2+-Fe = -0.441 V. Calculate the cell potential after the reaction has operated long enough for the Fe2+ to have changed by 0.399 mol/L.
Consider the cell described below at 253 K: Sn | Sn2+ (0.979 M) || Pb2+ (0.943 M) | Pb Given E°pb2+_Pb = -0.131 V, Eºsn2+_Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed 0.355 mol/L. .00893 V Submit Answer You have entered that answer before Incorrect. Tries 5/45 Previous Tries
HHelp ASAP
Consider the cell described below at 283 K: Sn | Sn2+ (0.871 M) || Pb2+ (0.927 M) | Pb Given E°Pb2+>Pb = -0.131 V, E°şn2+-Sn = -0.143 V. Calculate the cell potential after the reaction has operated long enough for the Sn2+ to have changed by 0.355 mol/L. -2780 V Incorrect. Tries 6/45 Previous Tries Submit Answer
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) e°(Fe2+ / Fe) = -0.4400 V, Eº (Cd2+ / Ca) = -0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+1 0.75 M and Ca2+] = 0.010 M. E- L v Will the reaction occur spontaneously at these conditions? O yes no cannot predict
Consider a galvanic cell based on the following two half reactions under standard conditions at 298 K: Fe2+ + 2 e → Fe -0.440 V Cd2+ + 2 e → Cd -0.403 V What will the potential of the cell be when the cathode solution concentration changes by 0.657 M? Cell potential =
Be sure to answer all parts. Consider the following reaction: Cd(s) + Fe2+(aq) → Cd2+(aq) + Fe(s) E° (Fe2+ / Fe) =-0.4400 V, Eº (Cd2+ / Cd) = -0.4000 V Calculate the emf for this reaction at 298 K if [Fe2+] = 0.85 M and [Cd2+] = 0.010 M. E= Lv Will the reaction occur spontaneously at these conditions? o o yes no cannot predict o
For the galvanic (voltaic) cell Cd2+ (aq) + Fe(s) Cd(s) + Fe2+(aq) (E° = 0.0400 V), what is the ratio [Fe2+1/[Cd2+] when E = 0.001 V? Assume T is 298 K 1 2. 3 Х 4 5 6 C 7 8 9 + +/- 0 x 100
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