Question

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)

Answer 1

a) ClO₃⁻(aq)

b) Cr₂O₇²⁻(aq)

Step-by-step 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.