Electrochemistry (Spontaneous Reactions)
For $\ce{Pb^2+_{(aq)} + Cu_{(s)}&lt;=&gt; Pb_{(s)} + Cu^2+_{(aq)}}$, your potentials are the wrong way around. The cathode is the electrode at which the reduction occurs so it's lead minus copper.

For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}&lt;=&gt; Ag_{(s)} + Br2_{(l)}}$, I don't know where you got .76 V for bromide, but that's not the right number. Also, consider the $K_{sp}$ of $\ce{AgBr}$ and think about whether a non-redox reaction is occurring.

For $\ce{3Li^+_{(aq)} + Cr_{(s)}&lt;=&gt; 3Li_{(s)} + Cr^3+_{(aq)}}$, your potentials are backwards again and the potential for $\ce{Cr}$ is wrong. Consider how easily elemental lithium is oxidized; just putting it in water will evolve hydrogen and oxidize the lithium, so anything that reduces the lithium cation would have to be a phenomenal reducing agent.

For $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}&lt;=&gt; 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, you have the right order, but the wrong potential for $\ce{Cd}$.

If you're having trouble keeping the order straight, it may help to think of it this way:
$$E_{\mathrm{cell}}=E_{\mathrm{reduction}} + E_{\mathrm{oxidation}}$$ where $E_{\mathrm{oxidation}}=-E_{\mathrm{reduction}}$. It's the same equation, but instead of keeping track of anode and cathode, the half-cell where reduction is happening will match what the table says and you use the potential as-is (since it's a table of reduction potentials), and the half-cell where oxidation is happening will be backwards from that in the table, so you negate the potential.