what is the cell voltage when saturated calomel and Pt electrodes are dipped into a solution containing 0.00217 M Br2 (aq) and 0.234 M Br^-
To determine the cell voltage when saturated calomel (Hg2Cl2) and Pt electrodes are dipped into a solution containing 0.00217 M Br2 (aq) and 0.234 M Br^-, we need to set up the balanced redox reaction and use the Nernst equation.
1. Write the balanced redox reaction:
2Br^- (aq) + Br2 (aq) ⇌ 3Br^-(aq)
2. Determine the half-reactions:
cathode: Br2 (aq) + 2e^- → 2Br^- (aq)
anode: 2Br^- (aq) → Br2 (aq) + 2e^-
3. Identify the standard electrode potentials:
E°(Br2/2Br^-) = 1.087 V (taken from reference tables)
4. Use the Nernst equation to calculate the cell voltage:
E(cell) = E°(cell) - (RT/nF) * ln(Q)
- R is the ideal gas constant (8.314 J/mol·K)
- T is the temperature in Kelvin
- n is the number of electrons transferred during the reaction
- F is the Faraday constant (96485 C/mol)
- Q is the reaction quotient
In this case, the cell voltage is at standard conditions (25°C or 298 K), so the Nernst equation becomes:
E(cell) = E°(cell) - (0.0592/n) * log(Q)
5. Calculate the cell voltage:
Since the concentration of Br2 and Br^- are given, we can use these concentrations to calculate Q:
Q = [Br^-]^3/[Br2] = (0.234 M)^3 / 0.00217 M = 32.39
Substituting the values into the Nernst equation:
E(cell) = 1.087 V - (0.0592/2) * log(32.39)
E(cell) = 1.087 V - 0.0298 * 1.510
E(cell) ≈ 1.04 V
Therefore, the cell voltage is approximately 1.04 V.