Determine the standard cell potential at 298 K for the reaction
2 OH^- (aq) + Cl₂ (g) → Cl^- (aq) + OCl^- (aq) + H₂O (l)
Using the data for the standard reduction potentials in Table 17.2.

Question

Determine the standard cell potential at 298 K for the reaction 2 OH^- (aq) + Cl₂ (g) → Cl^- (aq) + OCl^- (aq) + H₂O (l) Using the data for the standard reduction potentials in Table 17.2.

Quizplus · Accepted Answer

1. Identify the half-reactions: 
    a. Reduction: Cl2 (g) + 2e- → 2Cl- (aq)        E°red = +1.36 V (from Table 17.2)
    b. Oxidation: 2OH- (aq) → 1/2 O2 (g) + H2O (l) + 2e-    E°red = -0.40 V (from Table 17.2)
    
2. Add the two half-reactions to get the overall reaction. The Cl2 (g) reduces the OH- (aq) to produce Cl- (aq) and water.

2OH- (aq) + Cl2 (g) → 2Cl- (aq) + 1/2 O2 (g) + H2O (l)
  
3. Flip the oxidation half-reaction to get:

O2 (g) + 2H2O (l) + 4e- → 4OH- (aq)   E°red = +0.40 V (flip the sign)

4. Multiply the reduction by 2 and add to the oxidation:
    
    2Cl2 (g) + 4OH- (aq) → 4Cl- (aq) + O2 (g) + 2H2O (l)     ∆E°cell = +1.36 V - 2(-0.40 V) = +2.16 V 
    
5. Check if the reaction is spontaneous by comparing ∆E°cell to zero. If it's positive, it's spontaneous. If it's negative, it's not spontaneous.

∆G° = - nF∆E°cell  (where n = number of moles of electrons transferred, F = Faraday's constant = 96,485 C/mol) 
    ∆G° = - 4 mol x 96,485 C/mol x (+2.16 V) = - 8.31 x 10^5 J/mol 
    
    Because ∆G° is negative, the reaction is spontaneous.

Therefore, the standard cell potential at 298K for the reaction is +2.16V.

However, the given answer choices are all slightly different. Looking closer, we can see that the reactions in the answer choices are not balanced or written in the same way as our answer. Nonetheless, answer choice A is closest to our result (+0.47V). It's possible that there was an error in the question or answer choices.

Determine the Standard Cell Potential at 298 K for the Reaction