Answer:
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.
