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In the oxidation-reduction reaction

$\ce{Sn^{4+} + 2 Fe^{2+} -> 2 Fe^{3+} + Sn^{2+}}$

(A) $\ce{Sn^{4+}}$ is the oxidizing agent and $\ce{Fe^{2+}}$ is the reducing agent.

(B) $\ce{Sn^{4+}}$ is the reducing agent and $\ce{Fe^{2+}}$ is the oxidizing agent.

(C) $\ce{Sn^{4+}}$ is the reducing agent and $\ce{Fe^{3+}}$ is the oxidizing agent.

(D) $\ce{Fe^{3+}}$ is the oxidizing agent and $\ce{Sn^{2+}}$ is the reducing agent.

My thoughts: Because $\ce{Sn^{4+}}$ is getting reduced, and $\ce{Fe^{2+}}$ is getting oxidized, then $\ce{Sn^{4+}}$ is therefore an oxidizing agent, and $\ce{Fe^{2+}}$ is a reducing agent, which leads me to believe that A is the right answer. However, the answer key says D. I note that (D) is true for the reverse reaction, but why would A not be true then?

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    $\begingroup$ Yea, D lists the products not the reactants. Sometimes the answer key is wrong and A is right ;) $\endgroup$ – airhuff Apr 28 '17 at 4:11
  • $\begingroup$ If I were you @DisplayName, I'd avoid using that book. $\endgroup$ – Pritt Balagopal Apr 28 '17 at 4:38

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