The standard reduction potential for the reduction of RuO4 - (aq) to RuO42 - (aq) is + 0.59 V.

You may want to reference (Pages 860 - 867) Section 20.4 while completing this problem.

Use Appendix E in the textbook. Which of the following substances can oxidize RuO42 - (aq) to RuO4 - (aq) under standard conditions?

a) ClO3 - (aq)

b) Cr2O72 - (aq)

c) Ni2 + (aq)

d) Pb2 + (aq)

e) I2 (s)

a) ClO₃⁻ (aq)

b) Cr₂O₇²⁻ (aq)

Explanation:

The ability of a species to act as an oxidizing agent depends on the standard reduction potential (E°red). The higher the E°red, the more able it is to be an oxidizing agent.

We want to oxidize RuO₄²⁻ (aq) to RuO₄⁻ (aq). The inverse reduction has a E°red = + 0.59 V. Then, we need an species with a higher standard reduction potential.

Let's consider the following standard reduction potentials.

a) ClO₃⁻ (aq) / ClO₂ (g) E°red = 1.18 V

b) Cr₂O₇²⁻ (aq) / Cr³⁺ (aq) E°red = 1.33 V

c) Ni²⁺ (aq) / Ni (s) E°red = - 0.25 V

d) Pb²⁺ (aq) / Pb (s) E°red = - 0.13 V

e) I₂ (s) / I⁻ (aq) E°red = 0.54 V

ClO₃⁻ (aq) and Cr₂O₇²⁻ (aq) can oxidize RuO₄²⁻ (aq) under standard conditions.