Question

Write the balanced cell reaction for the cell schematic below, calculate the standard cell potential, and note whether the reaction is spontaneous under standard state conditions. Pt(s)│H₂(g)│H⁺(aq)║Br₂(aq), Br⁻(aq)│Pt(s)

a) Cell Reaction: 2H⁺(aq) + 2Br⁻(aq) + H₂(g) → 2HBr(aq), Standard Cell Potential: 1.23 V, Spontaneous: Yes
b) Cell Reaction: 2HBr(aq) → 2H⁺(aq) + 2Br⁻(aq) + H₂(g), Standard Cell Potential: -1.23 V, Spontaneous: No
c) Cell Reaction: H₂(g) + 2Br⁻(aq) → 2HBr(aq), Standard Cell Potential: 0.23 V, Spontaneous: Yes
d) Cell Reaction: 2HBr(aq) → H₂(g) + 2Br⁻(aq), Standard Cell Potential: -0.23 V, Spontaneous: No

Answer 1

The balanced cell reaction is H₂(g) + Br₂(aq) → 2H⁺(aq) + 2Br⁻(aq), and the standard cell potential is 1.09 V, indicating that the reaction is spontaneous under standard state conditions.

Step-by-step explanation:

To write the balanced cell reaction for the given electrochemical cell, we need to identify the half-reactions at the anode and cathode. In this case, we have hydrogen gas at the anode and bromine in the aqueous phase at the cathode. The half-reactions and their standard potentials (E°) are as follows:

• Anode (oxidation): H₂(g) → 2H⁺(aq) + 2e⁻ E° = 0.00 V
• Cathode (reduction): Br₂(aq) + 2e⁻ → 2Br⁻(aq) E° = 1.09 V

By combining these half-reactions, we obtain the total cell reaction:

H₂(g) + Br₂(aq) → 2H⁺(aq) + 2Br⁻(aq)

To calculate the standard cell potential, E°cell, we take the difference between the cathode and anode potentials:

E°cell = E°cathode - E°anode = 1.09 V - 0.00 V = 1.09 V

A positive standard cell potential indicates that the reaction is spontaneous under standard state conditions. Therefore, the reaction will proceed without the need for an external voltage.