Question

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

Answer 1

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.