Question

An electrochemical cell is based on the following two half-reactions: Oxidation: Pb(s) ? Pb2+(aq, 0.18 M) + 2e-, E° = -0.13 V Reduction: MnO4-(aq, 1.20 M) + 4H+(aq, 1.3 M) + 3e- ? MnO2(s) + 2H2O(l), E° = 1.68 V Part A Compute the cell potential at 25 °C. Express the cell potential in volts to three significant figures. Ecell = 1.85 V

          An electrochemical cell is based on the following two half-reactions:
Oxidation: Pb(s) ? Pb2+(aq, 0.18 M) + 2e-, E° = -0.13 V
Reduction: MnO4-(aq, 1.20 M) + 4H+(aq, 1.3 M) + 3e- ? MnO2(s) + 2H2O(l), E° = 1.68 V
Part A
Compute the cell potential at 25 °C.
Express the cell potential in volts to three significant figures.
Ecell = 1.85 V

An electrochemical cell is based on the following two half-reactions:
Oxidation: Pb(s) ? Pb2+(aq, 0.18 M) + 2e-, E° = -0.13 V
Reduction: MnO4-(aq, 1.20 M) + 4H+(aq, 1.3 M) + 3e- ? MnO2(s) + 2H2O(l), E° = 1.68 V
Part A
Compute the cell potential at 25 °C.
Express the cell potential in volts to three significant figures.
Ecell = 1.85 V

Added by Mark R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

05/12/2022

Step 1

18 M) + 2e- Reduction: MnO4-(aq; 1.20 M) + 4H+(aq; 1.3 M) + 3e- → MnO2(s) + 2H2O(l) ** Show more…

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An electrochemical cell is based on the following two half-reactions: Oxidation: Pb(s) → Pb2+(aq, 0.18 M) + 2e-, E° = -0.13 V Reduction: MnO4-(aq, 1.20 M) + 4H+(aq, 1.3 M) + 3e- → MnO2(s) + 2H2O(l), E° = 1.68 V Part A Compute the cell potential at 25 °C. Express the cell potential in volts to three significant figures.