We’re asked to determine the ∆G° at 25°C for the hall-reactions:
Cr2O72- + 14H+ + 6e-⇌ 2Cr3+ + 7H2O E° = 1.20 V
H2O2 + 2H+ + 2e-⇌ 2 H2O E° = 1.75 V
For this problem, follow the steps:
Step 1. Determine the cathode and anode and the number of electrons involved
Step 2. Calculate the Ecell
Step 3. Calculate for the ΔG
Step 1. Determine the oxidation and half reactions:
Loss of Electron is an Oxidation process undergone by the Reducing Agent (LEORA). In the Anode, Oxidation happens (An Ox)
Gain of Electron is a Reduction process undergone by the Oxidizing Agent (GEROA). Reduction happens in the Cathode (Red Cat)
You can determine the anode and cathode by comparing their E° values.
• ↓ E° → oxidation → anode
• ↑ E° → reduction → cathode
Consider a galvanic cell built on the following half-reactions with their standard reduction potentials shown. What is ∆G° at 25 °C for the reaction involved?
Cr2O72- + 14H+ + 6e- ⇌ 2Cr3+ + 7H2O E° = 1.20 V
H2O2 + 2H+ + 2e- ⇌ 2 H2O E° = 1.75 V
a. +106 kJ
b. -106 kJ
c. -318 kJ
d. 318 kJ
e. 0 kJ
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