When the Cu²+ concentration is 9.81x10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.428V. What is the Mn²+ concentration? Cu²+ (aq) + Mn(s)→→→→→ Cu(s) + Mn²+ (aq) Answer: M

When the Hg2+ concentration is 1.11×10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.916V. What is the Mn2+ concentration?Hg2+(aq) + Mn(s) Hg(l) + Mn2+(aq)

Payal

When the Pb2+ concentration is 5.46×10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.147V. What is the Mg2+ concentration?Pb2+(aq) + Mg(s) Pb(s) + Mg2+(aq)

Zn(s) │ Zn2+(aq, 0.10 M) ║ Zn2+(aq, 0.50 M) │ Zn(s) The concentration cell above was allowed to operate until the cell reaction reached equilibrium. What are the cell potential and the concentrations of zinc(II) in each half-cell for the cell now?

When the Hg2+ concentration is 1.63×104 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.401V. What is the A13+ concentration? 3Hg²+(aq) + 2Al(s)—>3Hg(1) + 2A13+(aq) Answer: M

When the Ag+ concentration is 4.19×10-4 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.257V. What is the Al3+ concentration? 3Ag+(aq) + Al(s)3Ag(s) + Al3+(aq)Answer: M

When the Hg+ concentration is 8.47×104 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 2.422V. What is the Al3+ concentration? 3Hg*"(aq) + 2Al(s)- →3H9(1) + 2A1*(aq) Answer: 7.53x10^6

Based on the cell potentials in the previous question calculate equilibrium constant of the following reaction                                                   H2(g) + I2 (g)  →  2HI(aq) Please enter your answer using E notation with 2 decimals.  for example, 1234.456 is written as 1.23E3

Using standard potentials given in the appendices, calculate the standard cell potentials for the following reaction: Cu(s) + 2Ag+ (aq) = Cu²+ + 2Ag(s) Report the answer in volts with three significant figures. 0.462 V Submit Previous Answers Correct Correct answer is shown. Your answer 0.46 V was either rounded differently or used a different number of significant figures than required for this part. Part B Determine the equilibrium constant for the above reaction.