We need at least 10 more requests to produce the answer.
0 / 10 have requested this problem solution
The more requests, the faster the answer.
What is the potential of a cell made up of Zn/Zn2+ and Cu/Cu2+ half-cells at 25°C...
What is the potential of a cell made up of Zn / Zn2+ and Cu / Cu2+ half-cells at 25°C if[Zn2+] = 0.31 M and [Cu2+] = 0.51 M ?
The voltage of the cell combination of Cu/Cu2+ and Zn/Zn2+ is 0.953V. The standard reduction potential for Cu/Cu2+ is expected to be 0.34V. Determine the reduction potential for Zn/Zn2+
A 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 1.00 M and 0.150 M respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.250 M? (Hint: What should the concentration of Zn2+ be if Ni2+ has fallen from its original concentration to its current concentration?)
For the following cell what is the cell potential under standard conditions? Zn + Cu2+ → Cu + Zn2+ If Cu2+ + 2e- → Cu E° = 0.34V and Zn2+ + 2e- → Zn E° = -0.76 V
What is E for a cell where E°=1.10 (Cu2++Zn→Cu+Zn2+), [Cu2+]=1.25 M, and [Zn2+]=0.0075 M at SATP? 1.25 V 0.95 V 1.03 V 1.17 V
A voltaic cell consists of a Pb/Pb2+ half-cell and a Cu/Cu2+ half-cell at 25 ∘C. The initial concentrations of Pb2+ and Cu2+ are 0.0530 M and 1.60 M , respectively. 1. What is the initial cell potential? 2. What is the cell potential when the concentration of Cu2+ has fallen to 0.200 M ? 3. What is the concentration of Pb2+ when the cell potential falls to 0.370 V ?
A 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 1.60molL−1 and 0.130 molL−1, respectively. Zn2+(aq)+2e−→Zn(s)E∘=−0.76V Ni2+(aq)+2e−→Ni(s)E∘=−0.23V Part A What is the initial cell potential? Part B What is the cell potential when the concentration of Ni2+ has fallen to 0.600 molL−1? Part C What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?
Choose the correct Nernst equation for the cell Zn + Cu2+ à Zn2+ + Cu E = E° - 0.0296 log(Cu / Zn) E=E° - 0.059 log([Zn2+] / [Cu2+]) E = E° - 0.0296 log(Zn / Cu) E = E° - 0.0296 log([Zn2+] / [Cu2+]) E = E° - 0.0296 log([Cu2+] / [Zn2+]) 20 points
What is E for a cell where Eº=1.10 (Cu2++Zn--Cu+Zn2+), [Cu2+3=1.50 M, and [Zn2+2=0.0050 M at SATP? 0.93 V 1.27 V 1.17 V 1.03 V
A 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 1.80 molL−1 and 0.130 molL−1, respectively. Zn2+(aq)+2e−→Zn(s)E∘=−0.76V Ni2+(aq)+2e−→Ni(s)E∘=−0.23V A) What is the cell potential when the concentration of Ni2+ has fallen to 0.600 molL−1? B) What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?