3L9 Standard reduction potentials for zinc(II) and copper(II)The standard reduction potential for a substance indicates how readily that substance gains electronsrelative to other substances at standard conditions. Consider the following:Zn2 2+(aq)+2e →Zn(s),Part BE cell =What is the standard potential, E cell, for this galvanic cell? Use the given standard reductionpotentials in your calculation as appropriate.Express your answer to three decimal places and include the appropriate units.► View Available Hint(s)Submit2+Cu²+ (aq) +2e¯→Cu(s),0OμAValueEred=-0.763 V= +0.337 VEred=Units?

The given question relates to E0cell E0cell = E0cathode - E0anodeE0 indicates standard reduction…

Answered: Standard reduction potentials for zinc…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter32: The Reception And Transmission Of Extracellular Information

Section: Chapter Questions

Problem 14P

See similar textbooks

Similar questions

Neutral sphingomyelinase 2 converts sphingomyelin into ceramide and phosphorcholine. What kind of enzyme is it? Assume Vmax is 35 µM min-1. When you provide 3.0 x 10-5 M of sphingomyelin, you observe an initial velocity of 6.0 µM min-1. Calculate the KM.
Calculate the concentration of an EDTA solution of which 24.22 mL were needed to titrate the Ca2+ions present in a 50.00 mL solution containing 246.7 mg CaCO3. Express your answer in terms of (a) molar concentration of EDTA (b) Ca Titer (mg Ca per mL of EDTA) MM Ca= 40.078 solve it in a way that it is in one equation already and cancelling all the values just like given in th epicture
SHOW COMPLETE SOLUTION 1. An electrochemical cell consist of 0.2715 Ibs of MnO2, and 0.2524 Ibs of PbO2. A current flow through a 1.1 ohm resistor of electrochemical cell for 45 mins. The reaction takes place at 400k and the total volume of the solution is 504.9 cm^3. Use the standard reduction cell potential table for MnO2 and PbO2: concentration of Mn2+ (2.81 M) and PbO2 (0.95 M) Cell Potential E=0.48 V current when it passes through a 1.1 ohm resistor = 0.44 ampere Determine the following: a. quantity of electricity/charge, Q b. Energy consumed in Joules c. Power in Kilowatts
Pharmaceuticals When a medicine is administered orally, the concentration of the drug in the patient's circulation after t minutes is given by C(t) = 0.06t – 0.00021, where Osls 240 and the concentration is measured in mg/L. When does the highest serum concentration occur, and what is that maximum serum concentration?
At Electrochemical Equilibrium, which best describes the distribution of ions in the chamber, on either side of the membrane, when one "mole" of salt (NaCl) is added to one side of a 1 liter chamber, filled with water and divided in half by a membrane permeable only to Na+. 2.0 M CI- and 1.25 M Na+ one side of the membrane; 0.75 M Na+ on the other side of the membrane. 1.0 M CI- and 1.0 M Na+ on both sides of the membrane. 2.0 M CI- and 1.0 M Na+ one side of the membrane; 1.0 M Na+ on the other side of the membrane. 2.0 M CI- and 2.0 M Na+ on one side of the membrane; pure water on the other side of the membrane.
A Br selective electrode is applied for measuring bromide ions in saline solution. Cl' ions were found to interfere, with a kPot = 0.0050. %3D After calibration, the ISE equation was determined to be y = -70 mV - 59.2mV*log(x). If we measure a potential of 93 mV, and the [CI]=0.100 M, what was the [Br'] in units of M? O 0.00821 O 0.100 O 0.0050 O 0.123 O Correct answer not found in other options
The E1% at 280 nm of the protein myosin is 0.842. You take 40µl of a myosin solution of unknown concentration and add water to a final volume of 2.0 ml. The absorbance of this solution at 280 nm is 0.771. What is the concentration of the undiluted myosin solution? (units = mg/ml).
Use the appropriate equation to calculate the free energy change for the movement of Nat into the depolarized nerve cell where the membrane potential is +50mV and the Nain/Naout ratio is 7.3/1. Assume the temperature is 37°C. Stop! For full credit: 1. Enter your answer in kJ, kilojoules. 2. Do not put units in the box. 3. Round your answer to the nearest tenth. If your calculator permits, do not round until the end for the sake of continuity. 4. If your answer is negative, include the sign and number with no space.
What would be the free-energy change generated by the import of one mole of Na+ from a concentration of 10 mM to 150 mM with a membrane potential of −25 mV at 37°C? Give your answer without units and to one decimal place. F = 96.5 kJ/V•mol
Theoretically, the pI of myoglobin is 7.2. In phosphate buffer pH 6.0, what is the overall charge of myoglobin? How would it interact with CM Sepharose?
The acidity of the stomach is maintained by the H*/K* ATPase in parietal cells of the gastric mucosa. These cells have an internal pH = 7.4. The H*/K* ATPase transports H* across the cellular membrane into the stomach where pH = 0.8. (a) Given a membrane potential of 65 mV (inside negative), calculate AG in kJ/mol for the transport of H* into the stomach at 37 °C. Show your work. (b) The hydrolysis of ATP is directly coupled to H* transport and the stoichiometry of the transport reaction is such that 1 mol of ATP is hydrolyzed for every 1 mol of H* transported into the stomach. ATP is hydrolyzed by the ATPase on the inside of the cell. Under the conditions given in part (a) and assuming the steady-state concentrations of ATP, ADP and P₁ given below, is the transport of H* in to the stomach thermodynamically favorable at 37 °C? Show the calculation that supports your answer. Steady-state intracellular concentrations: ATP = 4.2 mM ADP = 360 μM Pi = 5.4 mM For ATP hydrolysis at 37 °C: ATP…
a) (1) Calculate the physiological AG (not AG.) for the reaction: Phosphocreatine + ADP - creatine + ATP Given; Phosphocreatine + H;0 - creatine + Pi ADP + Pi → ATP + H;0 AG.' -43 kJ/mol AG.- +30.5 kJ/mol at 25°C as it occurs in the cytosol of neurons, in which phosphocreatine is present at 4.7 mM, creatine at 1.0 mM, ADP at 0.20 mM, and ATP at 2.6 mM. (R = 8.315 JK-' mol-) (ii) Caleulate the free energy change at standard conditions for the following reaction: Acetaldehyde + NADH + H* + Ethanol + NAD* The half- reactions are: Acetaldehyde + 2H + 2e + Ethanol E°- - 0.20V NAD-+ 2H- + 2e ++ NADH + H- E=-0.32V (F= 96.485 kJ/V/mol)

SEE MORE QUESTIONS

Recommended textbooks for you

Biochemistry

ISBN:

9781305577206

Author:

Reginald H. Garrett, Charles M. Grisham

Publisher:

Cengage Learning

Anatomy & Physiology

Biology

ISBN:

9781938168130

Author:

Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble

Publisher:

OpenStax College

Human Physiology: From Cells to Systems (MindTap …

Biology

ISBN:

9781285866932

Author:

Lauralee Sherwood

Publisher:

Cengage Learning

Biochemistry

ISBN:

9781305577206

Author:

Reginald H. Garrett, Charles M. Grisham

Publisher:

Cengage Learning

Anatomy & Physiology

Biology

ISBN:

9781938168130

Author:

Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble

Publisher:

OpenStax College

Human Physiology: From Cells to Systems (MindTap …

Biology

ISBN:

9781285866932

Author:

Lauralee Sherwood

Publisher:

Cengage Learning

SEE MORE TEXTBOOKS