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 |
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
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
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
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
Question 10 of 11 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 = 0.002 V? Assume T is 298 K (1 2 3
Calculate \(E_{\text {cell }}^{\circ}\) for a iron-cadmium cell in which the cell reaction is$$ 2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) $$$$ \begin{array}{ll} \mathrm{Fe}^{3+}+e^{-} \rightarrow \mathrm{Fe}^{2+} & E^{\mathrm{O}}=0.77 \mathrm{~V} \\ \mathrm{Cd}^{2+}+2 e^{-} \rightarrow \mathrm{Cd} & E^{\mathrm{O}}=-0.40 \mathrm{~V} \end{array} $$is oxidized in the reaction: \(2 \mathrm{Fe}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \rightarrow 2 \mathrm{Fe}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s})\)Select one:A. -0.37 VB. 0.37 VC. -1.17 VD. 1.17 VE. none of these
Consider the half‑reaction for the reduction of Cd 2+ Cd2+ to
Cd(s) Cd(s) . Cd 2 +(aq)+2 e − ⟶Cd(s) ? ∘ Cd 2+ /Cd =−0.403 V
Cd2+(aq)+2e−⟶Cd(s)ECd2+/Cd∘=−0.403 V Calculate the potential of the
cadmium electrode at 25 ∘ C 25 ∘C when immersed in a 0.0560 M
0.0560 M solution of CdBr 2 . CdBr2. ? Cd = ECd= V V Calculate the
potential of the cadmium electrode at 25 ∘ C 25 ∘C when immersed in
a 0.0330...
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.
Table 20.2 Half-reaction E (V) -0.74 Cr3+ (aq) 3e --Cr (s) Fe2+(aq)+2e-Fe (s) Fe3+ (aq) 0.440 +eFe2+ (s) Sn4+ (aq) 2eSn2 (aq) +0.771 +0.154 8. Based on Table 20.2, which of the following reactions will occur spontaneously as written? A) Sn4+ (aq)+ Fe3+ (aq) Sn2+ (aq) + Fe2+ (aq) B) 3Fe (s)+2Cr3+ (aq)2Cr (s)+3F 2+ (aq) C) Sn4+ (aq)+ Fe2+ (aq) Sn2+ (aq)+ Fe (s) D) 3Sn4+ (aq)+ 2Cr (s)- 2Cr3+ (aq) + 3Sn2+ (aq) E) 3Fe2+ (aq) Fe (s)+2Fe3+...
1)A voltaic cell operates at 298 K according to the following reaction: 4 Fe2+(aq) + O2 (g) + 4 H+(aq) → 4 Fe3+ (aq) + 2 H2O (l) What is the emf of this cell when [Fe2+] = 6.5908E-4 M, [Fe3+] = 0.699 M, pressure O2 = 0.540 atm and pH = 3.10? 2)A voltaic cell operates at 298 K according to the following reaction: 3 Fe2+(aq) → Fe (s) + 2 Fe3+ (aq) What is the emf of this...