Question

Use the standard reduction potential table to predict which reaction will occur spontaneously.

2Fe3+(aq) + 2I–(aq) mc001-1.jpg 2Fe2+(aq) + I2(aq)

3Na+(aq) + Cr(s) mc001-2.jpg 3Na(s) + Cr3+(aq)

Cu(s) + 2HCl(aq) mc001-3.jpg Cu2+ + 2Cl–(aq) + H2(g)
Reducing Agent

Answer 1

Use standard reduction potentials (V): Fe3+/Fe2+ = +0.77, I2/2I– = +0.535, Na+/Na = −2.71, Cr3+/Cr = −0.74, Cu2+/Cu = +0.34, H+/H2 = 0.00.

1) 2Fe3+ + 2I– → 2Fe2+ + I2
- Reduction at cathode: Fe3+ + e– → Fe2+ (E° = +0.77)
- Oxidation at anode: 2I– → I2 + 2e– (reverse of I2 + 2e– → 2I–, E° = +0.535)
- E°cell = 0.77 − 0.535 = +0.235 V → spontaneous as written.
- Reducing agent (species oxidized): I–.

2) 3Na+ + Cr(s) → 3Na(s) + Cr3+
- As written: Na+ reduced (E° = −2.71), Cr oxidized (reverse of Cr3+/Cr, E° = −0.74)
- E°cell = (−2.71) − (−0.74) = −1.97 V → non‑spontaneous as written.
- (The spontaneous direction would be Cr3+ + 3Na(s) → Cr(s) + 3Na+; Na metal would be the reducing agent in that spontaneous direction.)
- Reducing agent in the written (but nonspontaneous) reaction: Cr(s).

3) Cu(s) + 2HCl → Cu2+ + 2Cl– + H2(g)
- As written: H+ reduced (E° = 0.00), Cu oxidized (reverse of Cu2+/Cu, E° = +0.34)
- E°cell = 0.00 − 0.34 = −0.34 V → non‑spontaneous as written.
- Reducing agent in the written (but nonspontaneous) reaction: Cu(s).

Summary: Only reaction 1 is spontaneous; the reducing agent there is I–. Reactions 2 and 3 are not spontaneous as written.