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55 solutions

The Nernst Equation

Q. Use the tabulated half-cell potentials to calculate the equilibrium constant (K) for the following balanced redox reaction at 25°C. 3 I2(s) + 2 Fe(s) → 2 Fe3+(aq) + 6 I⁻(aq)

Solved • Aug 12, 2019

The Nernst Equation

Q. A galvanic cell uses the reaction 3 Ce^4+(aq) + Cr(s) ---> 3 Ce^3+(aq) + Cr3+(aq). For this cell, Eo = 2.35 V at 25 C. If the ionic concentrations in the cell are 0.100 M for Ce4+(aq), 2.00 M for Ce3+(aq) and 3.00 M for Cr3+(aq), find the emf (voltage) of this cell at 25 C

Solved • Aug 9, 2019

The Nernst Equation

Q. A galvanice cell uses this half reaction: MnO-(aq) -> Mn2+(aq) in an acidic solution. Knowing that the standard potential Eo for this half reaction is 1.51 V pH 2 and the concentrations are [MnO4-] = 1.0 M and [Mn2+] = 1.0x10^-3 M. Calculate the potential E of this half cell at 25 C.

Solved • Aug 9, 2019

The Nernst Equation

Q. Voltaic cell operates based on the following two half reactions at pH = 2: Cr2O72-(aq) --> Cr3+(aq) Eo = +1.33 V I2(s) --> I-(aq) Eo = +0.54 V If the concentrations in the cell are: 0.10 M for Cr2O72-(aq), 10 M for Cr3+(aq), and 0.20 M I-(aq), calculate Ecell for this cell.

Solved • Aug 8, 2019

The Nernst Equation

Q. What is the value of the equilibrium constant at 25 oC for the reaction between the pair: Sn(s) and Pb2+(aq) to give Pb(s) and Sn2+(aq) Use the reduction potential values for Sn2+(aq) of -0.14 V and for Pb2+(aq) of -0.13 V Give your answer using E-notation with NO decimal places (e.g., 2 x 10-2 would be 2E-2; and 2.12 x 10-2 would also be 2E-2.).

Solved • Aug 7, 2019

The Nernst Equation

Q. Based on the following electrochemical cell, what is the standard reduction potential of metal M et 298 K? (R= 8.314J /K • mol, F= 96500 C/rnol) Half-Reaction E (V) Fe2-(aq) + 2e- -0.44 Fe(s) -0.54 V +0.60 V -0.30 V +0.56 V -0.28 V

Solved • Aug 6, 2019

The Nernst Equation

Q. Calculate the voltage of the cell Ag(s) | AgBr (s), NaBr (aq, 1.0M) || CdCl2 (aq, 0.050 M) | Cd (s) AgBr (s) + e-  Ag (s) + Br - E° = 0.071 V Ag+ + e-  Ag (s) E° = 0.799 V Cd2+ + 2e-  Cd (s) E° = -0.402 V Ksp (AgBr(s) ) = 5.0 x 10-13

Solved • Jul 29, 2019