# Balancing reduction equations for galvanic cell

Problem: Compute the potential of the following galvanic cell: $$(\ce{Mg (s)} \mid \ce{Mg^2+ (aq)}\ (0.30\ \mathrm{M}) \mid \mid \ce{Sn(OH)4^2- (aq)}\ (0.10\ \mathrm{M}); \mathrm{pH} = 9 \mid \ce{Sn (s)})$$

Attempt at solution: I know we have to use the formula $$E_\text{cell} = E - \frac{0.0592}{n} \log Q$$ The anode is situated on the left; this means electrons are going away there. So we have for the oxidation half-reaction: $$\ce{Mg -> Mg^2+ + 2e-}$$ The potential for this is $E_1 = 2.36\ \mathrm{V}$. Then I’m not sure how to write my reduction half-reaction. I just wrote: $$\ce{Sn(OH)4^2- + 2e- -> Sn(s)}$$ and $E_2 = -0.91\ \mathrm{V}$. But this doesn’t seem correct, because if I add both reactions to get the sum, my charges aren’t balanced anymore. Also, I wonder what happens with the $\ce{(OH)4^2-}$ ? It’s not on the right side of the arrow anymore.

To construct reduction equations there are a set of steps you can follow.

For a balanced equation, the number of atoms of each element, and the total charge, must be the same on each side of the equation.

Firstly identify the main reactants and products which you have correctly done in this case as: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s)}$$

Under acidic conditions the only things we can add to balance the reaction are $\ce{H=}$, $\ce{H2O}$ and $\ce{e-}$. Under alkaline conditions we can add $\ce{OH-}$ instead of $\ce{H+}$. Usually, when dealing with alkaline reactions the best way of dealing with them is to treat them as acidic and then neutralize the acid later. However, in this instance, since the only reactant involves hydroxide ions, we can immediately balance the hydrogens and oxygens just by adding four hydroxide ions to the products: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4OH- (aq)}$$

Then balance the charges to get the overall equation: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

The alternative method which works for any alkaline reactions, which are sometimes less obvious than the example above, is to balance the reaction under acidic conditions and then neutralize the acid.

Since we only have one option for balancing the oxygens we add $\ce{H2O}$ first: $$\ce{[Sn(OH)4]^{2-}(aq) -> Sn(s) + 4H2O(l)}$$

Now balance the hydrogens: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) -> Sn(s) + 4H2O(l)}$$

Now the charges: $$\ce{[Sn(OH)4]^{2-}(aq) + 4H+(aq) + 2e- -> Sn(s) + 4H2O(l)}$$

If this were in acidic solution then this would be the final balanced equation. However, since we are in alkaline solution, we add $\ce{OH-}$ ions to cancel out the $\ce{H+}$, forming $\ce{H2O}$. $$\ce{[Sn(OH)4]^{2-}(aq) + 4H2O(l) + 2e- -> Sn(s) + 4H2O(l) + 4OH- (aq)}$$

Finally we can cancel the excess $\ce{H2O}$ to give: $$\ce{[Sn(OH)4]^{2-}(aq) + 2e- -> Sn(s) + 4OH- (aq)}$$

This is the same result as we got earlier by the quicker method.

• Thanks for the reply. So if we add everything we get $\ce{Mg + Sn(OH)_4^{2-} -> Mg^{2+} + Sn + 4OH-}$. I used $E_1 = -2.36$ V and $E_2 = -0.91$V. Then should I add these two? If I do $E_{kathode} - E_{anode}$ I get $1.45$ V. If I then use the Nernst equation with $\log(0.30/0.10)$ I get $1.43$ V as answer. But my textbook says it should be $2.03$ V? – Kamil Jun 3 '15 at 16:29
• I agree with you. I get 1.44V as my answer (1.44/1.43 depends on what values for constants you use and your rounding). Possibly an error in the book (not uncommon for textbook questions in my experience) or maybe someone else will spot something we haven't. – bon Jun 3 '15 at 17:02