Answer:
Cr₂O₇⁻²(aq) and ClO₃⁻(aq)
Explanation:
At a redox reaction, one substance must be reduced (gain electrons) and others must be oxidized (lose electrons). To evaluate the potential of the substance to be reduced, it's placed a reaction, in standards conditions, with H₂.
The potential reduction is quantified by E°, and as higher is the value of E°, as easy is to the compound to be reduced. So, at a redox reaction, the compound with the greatest E° will be reduced, and the other will be oxidized, in a spontaneous reaction. The values of E° are:
RuO₄⁻(aq) to RuO₄²⁻(aq) E° = + 0.59 V (the reduction reaction is the opposite of the oxidation reaction).
Ni⁺²(aq) E° = -0.257 V
I₂(s) E° = +0.535 V
Cr₂O₇⁻²(aq) E° = +1.33 V
ClO₃⁻(aq) E° = +0.890 V
Pb²⁺(aq) E° = -0.125 V
So, the substances that have E° higher than the E° of the RuO₄⁻²(aq) are Cr₂O₇⁻²(aq) and ClO₃⁻(aq), which are the substances that can oxidize RuO₄⁻(aq) to RuO₄²⁻(aq).
