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Redox and electrode potentials

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Question 3

The redox equilibria for a hydrogen–oxygen fuel cell in acidic solution are shown below.

2H+(aq)+2e−⇌H2(g)Eθ=0.00 V 2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g}) \quad E^\theta = 0.00\text{ V} 2H+(aq)+2e−⇌H2​(g)Eθ=0.00 V 12O2(g)+2H+(aq)+2e−⇌H2O(l)Eθ=+1.23 V \frac{1}{2}\text{O}_2(\text{g}) + 2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2\text{O}(\text{l}) \quad E^\theta = +1.23\text{ V} 21​O2​(g)+2H+(aq)+2e−⇌H2​O(l)Eθ=+1.23 V

What is the equation for the overall cell reaction?

H2O(l)→H2(g)+12O2(g)\text{H}_2\text{O}(\text{l}) \rightarrow \text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})H2​O(l)→H2​(g)+21​O2​(g)

4H+(aq)+12O2(g)→H2(g)+H2O(l)4\text{H}^+(\text{aq}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2(\text{g}) + \text{H}_2\text{O}(\text{l})4H+(aq)+21​O2​(g)→H2​(g)+H2​O(l)

H2(g)+H2O(l)→4H+(aq)+12O2(g)\text{H}_2(\text{g}) + \text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}^+(\text{aq}) + \frac{1}{2}\text{O}_2(\text{g})H2​(g)+H2​O(l)→4H+(aq)+21​O2​(g)

H2(g)+12O2(g)→H2O(l)\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})H2​(g)+21​O2​(g)→H2​O(l)

Redox and electrode potentials Questions

  1. A Level
  2. /Chemistry
  3. /Redox and electrode potentials