2 H2(g) + O2(g) → 2 H,0(1) ran = -474 kJ/mol AG Electrode Half-Reaction Standard Reduction Potentlal (V) 0,(g) + 4 H* (aq) + 4 e 2 H,0(1) 2 H*(aq) + 2 e- - H2(g) Cathode Anode 0.00 The operation of a hydrogen fuel cell under standard conditions relies on the chemical reaction represented above. The table provides the relevant reduction half-reactions and the standard reduction potentials. Based on the information given, which of the following equations can be used to calculate the standard reduction potential, in volts, of the half-reaction occurring at the cathode? E red (cathode) = 474 4x 06,500 E rea (cathode): -474,000 B 4x 06,500 () 96,500 E red (cathode) = 474 x96,500 D Ea(cathode) = 474.000 US O

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Chapter1: Chemical Foundations
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Briana So
18
20
21
241
30
«
2 H2(g) + O2(g) → 2 H20(1)
= -474 kJ/mol
AG
rzn
Electrode
Half-Reaction
Standard Reduction Potential (V)
Cathode
O2(g) + 4 H*(ag) + 4 e- 2 H20(1)
2 H*(ag) + 2 e
?
Anode
→ H2(g)
0.00
The operation of a hydrogen fuel cell under standard conditions relies on the chemical reaction represented above. The table provides the relevant reduction half-reactions and the standard
reduction potentials. Based on the information given, which of the following equations can be used to calculate the standard reduction potential, in volts, of the half-reaction occurring at the
cathode?
E pea(cathode) = (Tx 06,500
474
4xf
A
474,000
E rea (cathode)
4 x 96,500
4 x 96,500
E red (cathode)
474
4x 96.500
E rea (cathode) = -(4.000
D
US
Transcribed Image Text:Briana So 18 20 21 241 30 « 2 H2(g) + O2(g) → 2 H20(1) = -474 kJ/mol AG rzn Electrode Half-Reaction Standard Reduction Potential (V) Cathode O2(g) + 4 H*(ag) + 4 e- 2 H20(1) 2 H*(ag) + 2 e ? Anode → H2(g) 0.00 The operation of a hydrogen fuel cell under standard conditions relies on the chemical reaction represented above. The table provides the relevant reduction half-reactions and the standard reduction potentials. Based on the information given, which of the following equations can be used to calculate the standard reduction potential, in volts, of the half-reaction occurring at the cathode? E pea(cathode) = (Tx 06,500 474 4xf A 474,000 E rea (cathode) 4 x 96,500 4 x 96,500 E red (cathode) 474 4x 96.500 E rea (cathode) = -(4.000 D US
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