Electrochemistry (Spontaneous Reactions)

Question

Determine which of the following pairs of reactants will result in a spontaneous reaction at 25 celsius?

Pb2+(aq) + Cu(s)

Ag+(aq) + Br-(aq)

Li+(aq) + Cr(s)

Fe3+(aq) + Cd(s)

None of the Above.

I've tried using using tables like this http://ch302.cm.utexas.edu/images302/Electrochemistry_Reduction_Potentials.jpg to use the equation Ecell = Ecath - Eanode.

I'm hoping anyone could answer the question, and explain why each of them could be, or could not be spontaneous. (I know that an Ecell being positive results in spontaneity).

Here's my work:

Pb2+(aq) + Cu(s)

.34-(-.13) = .47

Ag+(aq) + Br-(aq)

.80 - (.76) = .04

Li+(aq) + Cr(s)

-.90-(-3.05) = 2.15

Fe3+(aq) + Cd(s)

-.04 - (-.81) = 77

None of the Above.

As you can see, they're all spontaneous. I'm wondering if I grabbed the wrong values from the above table. The question asks about ONE spontaneous reaction, from my calculations, they're all spontaneous.

@MichaelD.M.Dryden

Answer 1

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

For $\ce{2 Ag+_{(aq)} + 2 Br^{-}_{(aq)}<=> Ag_{(s)} + Br2_{(l)}}$, 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 $\ce{3Li^+_{(aq)} + Cr_{(s)}<=> 3Li_{(s)} + Cr^3+_{(aq)}}$, 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 $\ce{2Fe^3+_{(aq)} + 3Cd_{(s)}<=> 2Fe_{(s)} + 3Cd^2+_{(aq)}}$, 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.