11:27The Nernst equation | Applications of thermodynamics | AP Chemistry | Khan AcademyKhan Academy497views
02:32How to find the cell potential under nonstandard conditions| Nernst EquationMelissa Maribel461views
13:2719.5 How to Calculate Nonstandard Cell Potential [Nernst Equation] | General ChemistryChad's Prep337views
11:30Using the Nernst equation | Redox reactions and electrochemistry | Chemistry | Khan AcademyKhan Academy Organic Chemistry429views
30:53Nernst Equation Explained, Electrochemistry, Example Problems, pH, Chemistry, Galvanic CellThe Organic Chemistry Tutor596views
Multiple ChoiceIf [Br–] = 0.010 M and [Al3+] = 0.022 M, predict whether the following reaction would proceed spontaneously as written at 25ºC:Al (s) + Br2 (l) ⇌ Al3+ (aq) + Br– (aq)Standard Reduction PotentialsAl3+ (aq) + 3 e– → Al (s) E°red = –1.66 VBr2 (l) + 2 e– → Br– (aq) E°red = +1.09 V167views3rank
Multiple ChoiceDetermine [Fe2+] for the following galvanic cell at 25ºC if given [Sn2+] = 0.072 M, [Fe3+] = 0.0219 M, and [Sn4+] = 0.00345 M. Sn2+ (aq) + 2 Fe3+ (aq) ⇌. Sn4+ (aq) + 2 Fe2+ (aq) Ecell = + 0.68 VStandard Reduction PotentialsSn4+ (aq) + 2 e– →. Sn2+ (aq) E°red = + 0.151 VFe3+ (aq) + e– → Fe2+ (aq) E°red = + 0.771 V168views1comments
Textbook QuestionConsider the following voltaic cell: (c) What is the change in the cell voltage when the ion concentrations in the cathode half-cell are increased by a factor of 10?703views
Textbook QuestionConsider a Daniell cell with 1.0 M ion concentrations: Does the cell voltage increase, decrease, or remain the same when each of the following changes is made? Explain. (a) Write a balanced equation for each cell reaction.921views
Textbook QuestionA voltaic cell utilizes the following reaction: Al1s2 + 3 Ag+1aq2 ¡ Al3+1aq2 + 3 Ag1s2 What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode half-cell, diluting the solution.787views
Textbook QuestionA voltaic cell is constructed that uses the following reaction and operates at 298 K: Zn1s2 + Ni2+1aq2 ¡ Zn2+1aq2 + Ni1s2 (c) What is the emf of the cell when 3Ni2+4 = 0.200 M and 3Zn2+4 = 0.900 M?806views
Textbook QuestionA voltaic cell utilizes the following reaction: 4 Fe2+1aq2 + O21g2 + 4 H+1aq2 ¡ 4 Fe3+1aq2 + 2 H2O1l2 (b) What is the emf of this cell when 3Fe2+4 = 1.3 M, 3Fe3+4= 0.010 M, PO2 = 0.50 atm, and the pH of the solution in the cathode half-cell is 3.50?711views
Textbook QuestionA voltaic cell utilizes the following reaction: (b) What is the emf for this cell when 3Fe3+4 = 3.50 M, PH2= 0.95 atm, 3Fe2+4 = 0.0010 M, and the pH in both half-cells is 4.00?972views1rank1comments
Textbook QuestionA voltaic cell is constructed that is based on the following reaction: Sn2+1aq2 + Pb1s2 ¡ Sn1s2 + Pb2+1aq2 (a) If the concentration of Sn2+ in the cathode half-cell is 1.00 M and the cell generates an emf of +0.22 V, what is the concentration of Pb2+ in the anode half-cell?894views
Textbook QuestionA voltaic cell employs the following redox reaction: Sn2+(aq) + Mn(s) ¡ Sn(s) + Mn2+(aq) Calculate the cell potential at 25 °C under each set of conditions. c. [Sn2+] = 2.00 M; [Mn2+] = 0.0100 M3404views1rank
Textbook QuestionAn electrochemical cell is based on these two half-reactions: Ox: Pb(s) -> Pb2+ (aq, 0.10 M) + 2 e- Red: MnO4-(aq, 1.50 M) + 4 H+(aq, 2.0 M) + 3 e- -> MnO2(s) + 2 H2O(l) Calculate the cell potential at 25 °C.3232views
Textbook QuestionAn electrochemical cell is based on these two half-reactions: Ox: Sn(s) ¡ Sn2+(aq, 2.00 M) + 2 e- Red: ClO2(g, 0.100 atm) + e- ¡ ClO2-(aq, 2.00 M) Calculate the cell potential at 25 °C.2426views
Textbook QuestionA voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V? Ec°ell for the fuel-cell breathalyzer, which employs the following1195views
Textbook QuestionA voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M?2119views
Textbook QuestionA voltaic cell consists of a Zn>Zn2+ half-cell and a Ni>Ni2+ half-cell at 25 °C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential?557views
Textbook QuestionA voltaic cell consists of a Pb>Pb half-cell and a Cu>Cu half- cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. c. What are the concentrations of Pb2+ and Cu2+ when the cell potential falls to 0.35 V?779views
Textbook QuestionA voltaic cell consists of a Pb>Pb half-cell and a Cu>Cu half- cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. b. What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M?1504views
Textbook QuestionA voltaic cell consists of a Pb>Pb half-cell and a Cu>Cu half- cell at 25°C. The initial concentrations of Pb2+ and Cu2+ are 0.0500 M and 1.50 M, respectively. a. What is the initial cell potential?682views
Textbook QuestionA concentration cell consists of two Sn/Sn2+ half-cells. The cell has a potential of 0.10 V at 25°C. What is the ratio of the Sn2+ concentrations in the two half-cells?2331views2rank
Textbook QuestionA Cu/Cu2+ concentration cell has a voltage of 0.22 V at 25 °C. The concentration of Cu2+ in one of the half-cells is 1.5x10^-3 M. What is the concentration of Cu2+ in the other half-cell? (Assume the concentration in the unknown cell is the lower of the two concentrations.)3732views
Textbook QuestionConsider a galvanic cell that uses the reaction Calculate the potential at 25 °C for a cell that has the following ion concentrations: [Ag+] = 0.010M, [Ni2+] = 0.100M.1154views
Textbook QuestionA galvanic cell has an iron electrode in contact with 0.10 M FeSO4 and a copper electrode in contact with a CuSO4 solu-tion. If the measured cell potential at 25 °C is 0.67 V, what is the concentration of Cu2+ in the CuSO4 solution?1279views
Textbook QuestionThe Nernst equation applies to both cell reactions and half-reactions. For the conditions specified, calculate the potential for the following half-reactions at 25 °C. (b) 1215views1rank
Textbook QuestionWhen suspected drunk drivers are tested with a Breathalyzer, the alcohol (ethanol) in the exhaled breath is oxidized to acetic acid with an acidic solution of potassium dichromate: The color of the solution changes because some of the orange Cr2O7 2-is converted to the green Cr3+ The Breath-alyzer measures the color change and produces a meter reading calibrated in blood alcohol content. (b) What is the value of E for the reaction when the concentrations of ethanol, acetic acid, Cr2O7 are 1.0 M and the pH is 4.00?904views
Textbook QuestionA voltaic cell is constructed that uses the following half-cell reactions: Cu+1aq2 + e- ¡ Cu1s2 I21s2 + 2 e- ¡ 2 I-1aq2 The cell is operated at 298 K with 3Cu+4 = 0.25 M and 3I-4 = 0.035 M. (a) Determine E for the cell at these concentrations.2676views1rank
Textbook QuestionAt one time on Earth, iron was present mostly as iron(II). Later, once plants had produced a significant quantity of oxygen in the atmosphere, the iron became oxidized to iron(III). Show that Fe2+(aq) can be spontaneously oxidized to Fe3+(aq) by O2(g) at 25°C assuming the following reasonable environmental conditions: 709views
Textbook QuestionCopper reduces dilute nitric acid to nitric oxide (NO) but reduces concentrated nitric acid to nitrogen dioxide (NO2): Assuming that [Cu2+] = 0.10 M and that the partial pressures of NO and NO2 are 1.0 * 10-3 atm, calculate the potential (E) for reactions (1) and (2) at 25 °C and show which reaction has the greater thermodynamic tendency to occur when the concentration of HNO3 is (a) 1.0 M790views
Textbook QuestionA battery relies on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0 * 10-4 M and 1.5 M, respectively, in 1.0-liter half-cells. a. What is the initial voltage of the battery?1394views
Textbook QuestionA battery relies on the oxidation of magnesium and the reduction of Cu2+. The initial concentrations of Mg2+ and Cu2+ are 1.0 * 10-4 M and 1.5 M, respectively, in 1.0-liter half-cells. b. What is the voltage of the battery after delivering 5.0 A for 8.0 h?387views1rank
Textbook QuestionA mercury battery uses the following electrode half-reactions: (c) What is the effect on the cell voltage of a tenfold change in the concentration of KOH in the electrolyte? Explain..417views
Textbook QuestionAn MnO2(s)/Mn2+(aq) electrode in which the pH si 10.24 is prepared. Find the [Mn2+] necessary to lower the potential of the half-cell to 0.00 V (at 25°C)829views
Textbook QuestionTo what pH should you adjust a standard hydrogen electrode to get an electrode potential of -0.122 V? (Assume that the partial pressure of hydrogen gas remains at 1 atm.)653views
Textbook QuestionA concentration cell has the same half-reactions at the anode and cathode, but a voltage results from different concentrations in the two electrode compartments. (b) A similar cell has 0.10 M Cu2+ in both compartments. When a stoichiometric amount of ethylenediamine (NH2CH2CH2NH2) is added to one compartment, the measured cell potential is 0.179 V. Calculate the formation constant Kf for the complex ion Cu(NH2CH2CH2CH2)22+. Assume there is no volume change.739views
Textbook QuestionConsider the redox titration of 100.0 mL of a solution of 0.010 M Fe2+ in 1.50 M H2SO4 with a 0.010 M solution of KMnO4, yielding Fe3+ and Mn2+. The titration is carried out in an electrochemical cell equipped with a platinum electrode and a calomel reference electrode consisting of an Hg2Cl2/Hg electrode in contract with a saturated KCl solution having [Cl-] = 2.9M. Using any data in Appendixes C and D, calculate the cell potential after addition of (a) 5.0 mL, (b) 10.0mL, (c) 19.0 mL, and (d) 21.0 mL of the KMnO4 solution.386views