Question

Consider the following half-reactions. Which of these is the strongest reducing agent listed here?
I₂(s) + 2 e^- → 2 I^-(aq) E^o = 0.53 V
S₂O₈^2-(aq) + 2 e^- → 2 SO₄^2-(aq) E^o = 2.01 V
Cr₂O₇^2-(aq) + 14 H⁺ + 6 e^- → Cr³⁺(aq) + 7 H₂O(l) E^o = 1.33 V

	1.	I2(s)
	2.	I-(aq)
	3.	S2O82-(aq)
	4.	SO42-
	5.	Cr2O72-
	6.	Cr3+(aq)

Answer 1

I₂(s) + 2 e^- → 2 I^-(aq) E^o = 0.53 V  
S₂O₈^2-(aq) + 2 e^- → 2 SO₄^2-(aq) E^o = 2.01 V
Cr₂O₇^2-(aq) + 14 H⁺ + 6 e^- → Cr³⁺(aq) + 7 H₂O(l) E^o = 1.33 V

All the equations are reduction equation so the given values of electrode potential are standard reduction potential.

Lower the reduction potential value, better the reducing agent

So from the given values

I₂(s) + 2 e^- → 2 I^-(aq) E^o = 0.53 V    this is lowest

Hence, Iodine (I^-) is the best reducing agent