Examine the following reduction half-reactions and select the strongest oxidizin

Question

Examine the following reduction half-reactions and select the strongest oxidizing agent: Cr^2+ (aq) + 2e^- rightarrow Cr (s); E degree = - 0.91 V Co^2+ (aq) + 2e^- rightarrow Co(s); E degree = - 0.28 V 1 Fe^2+ (aq) + 2e^- rightarrow Fe (s); E degree = - 0.45 V Sr^2+ (aq) + 2e^- rightarrow Sr (s); E degree = - 2.7 V Pb^2+ (aq) + 2e^- rightarrow Pb(s); E degree = - 0.13 V Sr^2+ Pb^2+ Co^2+ Fe^2+ Cr^2+ The standard cell potential, E degree cell, for a voltaic cell formed for the combination of the following half - reactions is equal to [X] V. (Finn in the blank; do not show units.) Sn^4+ (aq) + 2e^- rightarrow Sn^2+ (aq); E degree cell = + 0.15 V Ni^2+ (aq) + 2e^- rightarrow Ni(s); E degree cell = + 0.28 V

Explanation / Answer

3) The strongest oxidizing agent is the compound which gets easily reduced

The lowest Potential is for the conversion of Pb(+2) to Pb(s), hence Pb can be easily reduced and act as the stronger oxidizing agent

The correct answer is Pb(+2), Option 2 is the correct answer

4)

Oxidation Half

Ni(s) --------- Ni(+2) + 2e- Eocell = 0.28V

Reduction Half

Sn(4+) + 2e- --------- Sn(+2) Eocell = 0.15V

Ecell = Eo oxidation + Eo reduction = 0.28 + 0.15 = 0.43V

Hence the correct answer is 0.43V

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