Use the standard reduction potentials below to determine which element or ion is

Question

Use the standard reduction potentials below to determine which element or ion is the best reducing agent. Pd2+(aq) + 2 e- rightarrow Pd(s) E degree = +0.90 V Sn2+(aq) + 2e rightarrow Sn(s) E degree = -0.14 V Cr2+(aq) + 2 e - rightarrow Cr(s) E degree = -0.91 V Pd2+ Pd(s) Sn2+ Cr2+ Cr(s) Which of the following species arc likely to behave as oxidizing agents: K+, MnO-4, Cr2O2-7, and I-? K+ only MnO-4 and Cr2O2-7 MnO-4, Cr202-7, and I- Cr2O2-7 and I- I- only Consider the following half-reactions: Cu2+(aq) + 2 e- rightarrow Cu(s) E degree = +0.34 V Sn2+(aq) + 2 e- rightarrow Sn(s) E degree = -0.14 V Fe2+(aq) + 2 e- rightarrow Fe(s) E degree = -0.44 V Al3+(aq) + 3 e- rightarrow Al(s) E degree = -1.66 V Mg2+(aq) + 2 e- rightarrow Mg(s) E degree = -2.37 V Which of the above metals or metal ions will reduce Fe2+(aq)? Cu(s) and Sn(s) Cu2+(aq) and Sn2+(aq) Al3+(aq) and Mg2+(aq) Al(s) and Mg(s) Sn(s) and Al3+(aq) Consider the following half-reactions: F2(g) + 2 e- rightarrow 2 F-(aq) E degree = +2.87 V Cu2+(aq) + 2 e- rightarrow Cu(s) E degree = +0.34 V Sn2+(aq) + 2 e- rightarrow Sn(s) E degree = -0.14 V Al3+(aq) + 3 e- rightarrow Al(s) E degree = -1.66 V Na+(aq) + e- rightarrow Na(s) E degree = -2.71 V Which of the above elements or ions will oxidize Sn(s)? Cu(s) and Al3+(aq) F-(aq) and Cu(s) F2(g) and Cu2+(aq) Al(s) and Na(s) Al3+(aq) and Na+(aq)

Explanation / Answer

16) option d, As Cr2+ has the lowest E0 value, as lower the E0 value more is the reducing ability.

17) option b, For an oxidising agent, it itself gets reduced having reduction in oxidation number, which is possible for MnO4- (+7), Cr2O7- (+6,+7)

18) option c, Fe2+ can only be reduced if E0 value is lower than its E0 value, ehich is in case for Al3+ and Mg2+.

19) option c, Sn (s) can be oxidised by E0 value more than its own as is in the case of F2 and Cu2+

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