What is E°_cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe³⁺(aq) + 2H₂O(l) → H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq)
Given:

Question

What is E°_cell for the following reaction, and is it spontaneous or nonspontaneous under standard-state conditions? 2Fe³⁺(aq) + 2H₂O(l) → H₂O₂(aq) + 2H⁺(aq) + 2Fe²⁺(aq) Given:

Quizplus · Accepted Answer

To calculate E°S1U1B1cellS1U1B0 for the given reaction, we need to use the standard reduction potentials of each half-reaction and plug them into the equation: 
E°S1U1B1cellS1U1B0 = E°S1U1B1(cathode) - E°S1U1B1(anode)
Here are the half-reactions and their respective standard reduction potentials:

FeS1U1P13+S1S1P0(aq) + 2e- → 2FeS1U1P12+S1S1P0(aq)    E°S1U1B1 = -0.22 V 
2H+(aq) + 2e- → H2(g)                                   E°S1U1B1 = 0 V 
O2(g) + 4H+(aq) + 4e- → 2H2O(l)                          E°S1U1B1 = 1.23 V
HS1U1B12S1U1B0O(l) + 2H+(aq) + 2e- → HS1U1B12S1U1B0S1U1B12S1U1B0(aq)  E°S1U1B1 = 0.68 V

Next, we need to balance the half-reactions and multiply them by coefficients so that the electrons cancel out. 
Half-reaction 1 (reduced form): 
2FeS1U1P13+S1S1P0(aq) + 2e- → 2FeS1U1P12+S1S1P0(aq)

Half-reaction 2 (oxidized form): 
2H+(aq) + 2FeS1U1P12+S1S1P0(aq) → H2S1U1P12+2+S1S1P0(aq) + 2FeS1U1P13+S1S1P0(aq)

Now we add the two half-reactions together and cancel out the spectator ions: 
2FeS1U1P13+S1S1P0(aq) + 2H+(aq) + O2(g) → H2S1U1P12+2+S1S1P0(aq) + 2FeS1U1P12+S1S1P0(aq)

Now we can calculate the overall standard cell potential: 
E°S1U1B1cellS1U1B0 = E°S1U1B1(cathode) - E°S1U1B1(anode)
E°S1U1B1cellS1U1B0 = (0.68 V + 0 V) - (-0.22 V + 1.23 V) = -1.00 V

Since the overall standard cell potential is negative, the reaction is nonspontaneous under standard-state conditions. Therefore, the best choice is (A).

What Is E°cell for the Following Reaction, and Is It

What Is E°_cell for the Following Reaction, and Is It