Determine products of redox-reaction (KMnO4) by Standard electrode potential
What you have listed are standard redox potentials. You need two of them to determine a reaction. As I'm not sure what reaction you intend, I'll give an example below, which you can use to work out your own reaction. The diagram helps to sort out what can happen.



Note that a reducing agent is an electron donor, and an oxidising agent an electron acceptor and $$\ce{oxidised form + electrons &lt;=&gt; reduced form}.$$

For example:
The oxidiser $\ce{Sn^{2+}}$ reacts completely with reducer $\ce{Cr^{2+}}$ (at standard redox potentials).
\begin{align}
E^\circ &amp;= \pu{-0.136 V}:&amp;
\ce{Sn^2+ + 2e- &amp;&lt;=&gt; Sn}\\ 
E^\circ &amp;= \pu{-0.41 V}:&amp;
\ce{Cr^3+ + e^- &amp;&lt;=&gt; Cr^2+}\\
&amp;&amp;
\ce{2Cr^2+ + Sn^2+  &amp;&lt;=&gt;&gt; 2Cr^3+ + Sn v}
\end{align}
With energy 
$$E= (\pu{-0.136 V}) - (\pu{-0.41 V}) = \pu{0.274 V}$$
 which corresponds to a negative free energy and hence is spontaneous: 
\begin{align}
\Delta G &amp;= -nFE= (-2\times  96485\times 0.274) \pu{J};&amp; 
(\pu{1J} &amp;= \pu{1C*V}).
\end{align}

Although the reverse reaction, $\ce{Cr^{3+}}$ to oxidise $\ce{Sn}$ metal could occur, this happens to a negligible extent.