Problem: Use tabulated half-cell potentials to calculate ΔG°rxn for each of the following reactions at 25 °C.O2(g) + 2H2O(l) + 2Cu(s) → 4OH−(aq) + 2Cu2+(aq) Express the energy change in kilojoules to one significant figure.

We are asked to calculate ΔG°rxn for each of the following reactions at 25 °C.

O₂(g) + 2H₂O(l) + 2Cu(s) → 4OH⁻(aq) + 2Cu²⁺(aq)

The relationship between ΔG and E_cell is

$\boxed{\mathbf{∆}\mathbf{G}\mathbf{=}\mathbf{-}{\mathbf{nFE}}_{\mathbf{cell}}}$

Step 1. Write the two half-cell reactions

O₂(g) + 2 H₂O(l) → 4OH⁻(aq)

Cu(s) → 2Cu²⁺(aq)

Step 2. Identify the oxidation half-reaction (anode) and the reduction half-reaction (cathode)

When a balanced reaction is given, identify the anode and the cathode half-reactions by determining the changes in oxidation states of each species based on the given reaction:

Recall:

Lose               Gain
Electron         Electrons
Oxidation       Reduction

cathode → reduction → oxidation number decreases

anode → oxidation → oxidation number increases

For ions: charge of the ion = oxidation number
For neutral atoms/compound: oxidation number = 0

Cu(s) →  2Cu²⁺(aq)
• oxidation number increased → lost electrons → oxidized → anode

O₂(g) + 2 H₂O(l) → 4OH⁻(aq)
• oxidation number decreased → reduced → cathode

Step 3. Determine the half-cell potentials (refer to the Standard Reduction Potential Table)

2 Cu(s) → 4 e^- + 2Cu²⁺(aq)                                        E°_cell = 0.342 V

O₂(g) + 2 H₂O(l) + 4 e^-→ 4OH⁻(aq)                            E°_cell = 0.401 V

Step 4. Calculate E°_cell.

$$\begin{gathered} \boxed{\mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\mathbf{E}{\mathbf{°}}_{\mathbf{cathode}}\mathbf{-}\mathbf{E}{\mathbf{°}}_{\mathbf{anode}}} \\ \mathbf{E}{\mathbf{°}}_{\mathbf{cell}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{401}\mathbf{ }\mathbf{V}\mathbf{ }\mathbf{-}\left(\mathbf{0}\mathbf{.}\mathbf{342}\mathbf{ }\mathbf{V}\right) \end{gathered}$$  

  E°_cell = 0.059 V