Examine the following half-reactions and select the strongest oxidizing agent am

ID: 1021364 • Letter: E

Question

Examine the following half-reactions and select the strongest oxidizing agent among the substances.

[PtCl4] 2– (aq) + 2e– Pt(s) + 4Cl– (aq) E° = 0.755 V

RuO4(s) + 8H+ (aq) + 8e– Ru(s) + 4H2O(l) E° = 1.038 V

FeO4 2– (aq) + 8H+ (aq) + 3e– Fe3+(aq) + 4H2O(l) E° = 2.07 V

H4XeO6(aq) + 2H+ (aq) + 2e– XeO3(aq) + 3H2O(l) E° = 2.42 V

A) [PtCl4] 2– (aq) B) RuO4(s) C) HFeO4 – (aq) D) H4XeO6(aq) E) Cl– (aq)

The study guide says the answer is D) H4XeO6(aq) but isn't this one being oxidized? So how can it be the strongest oxidizing agent?

Please help!

As per the given data

Each metal ions is undergoing reduction.

Pt+2 to Pt(0)

Ru+8 to Ru(0)

Fe+6 to Fe+3

Xe+8 to Xe+6

So in all cases we are provided with the reduction potential values ( Volts)

Remember: Higher the value of reduction potential better the oxidizing agent.

so out of these given examples the reduction potential of

H4XeO6(aq) + 2H+ (aq) + 2e– XeO3(aq) + 3H2O(l) E° = 2.42 V is highest

So H4XeO6(aq) will be best oxidizing agent.

[PtCl4] 2– will be best reducing agent

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