If [I-] = 0.050 M, calculate the half-cell potential for the following reaction: I2 (s) + 2e- ↔ 2I- E° = +0.540 V
If [I-] = 0.050 M, calculate the half-cell potential for the following reaction: I2 (s) +...
Use the standard half-cell potentials given, calculate the initial cell potential for the following reaction occurring in an electrochemical cell at 25°C when the starting concentrations of [Sn2+] = 2.0 M and [I2] = 0.010 M (The equation is balanced.) <Missing data [I−]=1.0M> I2 + Sn --> 2I- + Sn^2+ I2 + 2e- --> 2I- E=+0.54V Sn^2+ +2e- --> Sn E= -0.14V A) +0.47V B) +0.068V C) +0.75V D) +0.61V
For the reaction I2(s) + Cu(s) ↔ Cu2+(aq) + 2I-(aq) The following electrochemical cell is made: Cu(s)|Cu2+(aq, 0.1 M)||I‐(aq, 0.1 M)|I2(s)|C(s) At 25 C the cell potential for the above cell was measured and found to be 0.279 V. Calculate E°cel
1) Calculate the cell potential, E°cell, for the following electrochemical reaction. Cu(s) + I2(g) → Cu+2(aq) + 2I-1(aq) E = ? a.-0.87 V b.-0.19 V c.0.19 V d.0.87 V 2) Calculate the equilibrium constant (Keq) for the following reaction at 25 °C? (Faraday's Constant = 96,500 J / V · mol) & (R = 8.314 J / mol · K) 2Fe+3(aq) + Sn+2(aq) → 2Fe+2(aq) + Sn+4(aq) K = ? a. 4.04 x 107 b. 3.44 x 1018 c. 4.48...
An electrochemical cell is expressed as Cu(s) | Cu^2+ (0.20 M) || I^- (0.10 M) | I_3^- (0.20 M) | Pt. I_3^- + 2e^- rightarrow 3I^-, E^0 = 0.535 V Cu^2+ + 2e^- rightarrow Cu(s) E^0 = 0.339 V Please answer the following questions: a. Write half-cell reactions at anode and at cathode. b. Write whole-cell reaction. c. Calculate the potential/voltage of the cell. d. Calculate equilibrium constant for this whole-cell reaction.
Based on the half reactions given below, what is the E°cell, in volts, for the reaction performed in Experiment 22 (2Fe3+(aq) + 2I-(aq) ⟶ 2Fe2+(aq) + I2(aq))? Enter your answer to the thousandths place. Fe3+(aq) + e- ⟶ Fe2+(aq) E°red = 0.771 V I2(aq) + 2e- ⟶ 2I-(aq) E°red = 0.621 V
A chemist designs a galvanic cell that uses these two half-reactions: standard reduction potential half-reaction + O2(9)+4 H (aq)+4e' 2H20) = 1.23 V red Ered Fe+. (аq) Fe3(aq)+e = +0.771 V Answer the following questions about thiss cell Write a balanced equation for the half-reaction that happens at the cathode Write a balanced equation for the half-reaction that happens at the anode Write a balanced equation for the overall reaction that powers the cell. Be sure the reaction is spontaneous...
The standard cell potential (E°cell) of the reaction below is -0.55 V. The value of Δ G° for the reaction is ________ J/mol. I2 (s) + 2Br- (aq) → 2I- (aq) + Br2 (l) The standard cell potential (E°cell) of the reaction below is -0.55 V. The value of G° for the reaction is ________ J/mol. I2 (s) + 2Br- (aq) 2I- (aq) + Br2 (l) 5.5 ⋅ 10-6 1.1 ⋅ 105 0.55 0.54 none of the above
2. Using the information provided, calculate the standard cell potential, Eºcell, for the reaction below: Half-reaction Cr3+ (aq) + 3e- → Cr(s) Fe2+ (aq) + 2e- → Fe(s) Fe3+ (aq) + e- + Fe2+ (s) Sn4+ (aq) + 2e- + Sn2+ (aq) E° (V) -0.74 -0.440 +0.771 +0.154 3Sn(aq) +2Cr(s) →2Cr" (aq) +35n²+ (aq)
A galvanic cell is prepared using the following half-cells: (i) MnO4-(0.283M), Mn2+(0.081M), and H+(1.0 M), and (ii) Sn2+(0.236M) and Sn4+(0.145M). MnO4-+ 8H++ 5e-↔Mn2++ 4H2O Eo= 1.507 V Sn4++ 2e-↔Sn2+Eo= 0.151 V a)Calculate the galvanic cell voltage. b)Write the complete cell description. c)Write the cell reaction indicating clearly the direction of spontaneous reaction. d)Name the electrode (anode or cathode), give the polarity (positive or negative) of the electrode, and state the reaction (oxidation or reduction) occurring in the tin half-cell. Please...
Half-reaction E° (V) I2(s) + 2e- 2I-(aq) 0.535V Pb2+(aq) + 2e- Pb(s) -0.126V Cr3+(aq) + 3e- Cr(s) -0.740V The strongest oxidizing agent is: ______enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will I2(s) reduce Cr3+(aq) to Cr(s)? Which species can be reduced by Pb(s)? If none, leave box blank.