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Three electrolytic cells are joined in series to a battery, as the figure shows, so the same…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Three electrolytic cells are joined in series to a battery, as the figure shows, so the same current flows through each. Cell A holds copper(II) sulfate solution with two copper electrodes. Cell B holds $200\ \text{mL}$ of silver nitrate solution with $[\text{Ag}^+] = 0.0600\ \text{mol L}^{-1}$ and two platinum electrodes; it has no acid in it at the start. Cell C holds dilute sulfuric acid with two platinum electrodes. A steady current of $0.600\ \text{A}$ flows for $50.0\ \text{min}$ . At the cathode of cell B, silver ions are reduced in preference to anything else for as long as any are left. Oxygen forms at the platinum anodes. Take $F = 96487\ \text{C mol}^{-1}$ , molar masses $\text{Ag} = 107.9\ \text{g mol}^{-1}$ and $\text{Cu} = 63.5\ \text{g mol}^{-1}$ , and $R = 0.08314\ \text{L bar K}^{-1}\ \text{mol}^{-1}$ . Gas volumes are for the dry gases at $298\ \text{K}$ and $1\ \text{bar}$ . The volume of solution in cell B stays at $200\ \text{mL}$ .

Three electrolytic cells in a row, joined in series by wires to a battery and an ammeter that reads 0.600 A. Cell A holds copper(II) sulfate solution with two copper electrodes. Cell B holds 200 mL of silver nitrate solution, 0.0600 mol per litre, with two platinum electrodes. Cell C holds dilute sulfuric acid with two platinum electrodes. In each cell the electrode joined to the negative terminal is marked as the cathode.

How much charge passes?
What mass of silver is deposited in cell B?
What volume of hydrogen forms in cell B?
What mass does the copper anode of cell A lose?
What total volume of gas does cell C give?
What is the pH of the solution in cell B at the end?

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Plan. Count the electrons, then share them out in each cell. Cell B changes its cathode reaction partway through.

(a) Charge. $Q = It = 0.600 \times 50.0 \times 60\ \text{C} = 1800\ \text{C}$ , which is $\dfrac{1800}{96487} = 0.01866\ \text{mol}$ of electrons.

(b) Silver. Cell B has only $0.0600 \times 0.200\ \text{mol} = 0.0120\ \text{mol}$ of silver ions, fewer than the electrons. All of it deposits: $0.0120 \times 107.9\ \text{g} = 1.295\ \text{g}$ . The silver runs out after $32.2\ \text{min}$ .

(c) Hydrogen in cell B. The other $0.00666\ \text{mol}$ of electrons reduce $\text{H}^+$ : $2\text{H}^+ + 2e^- \rightarrow \text{H}_2$ . That is $0.00333\ \text{mol}$ of hydrogen, and $V = nRT/p = 0.00333 \times 24.776\ \text{L} = 82.4\ \text{mL}$ .

(d) Copper anode. It loses $\dfrac{0.01866}{2} \times 63.5\ \text{g} = 0.592\ \text{g}$ . The cathode gains the same mass, so the copper sulfate solution keeps its strength.

(e) Cell C. Water is split: hydrogen $\dfrac{0.01866}{2}$ and oxygen $\dfrac{0.01866}{4}$ , together $0.01399\ \text{mol}$ , so $V = 0.01399 \times 24.776\ \text{L} = 346.7\ \text{mL}$ .

(f) pH of cell B. The anode reaction $2\text{H}_2\text{O} \rightarrow \text{O}_2 + 4\text{H}^+ + 4e^-$ makes $0.01866\ \text{mol}$ of $\text{H}^+$ . The hydrogen at the cathode used $0.00666\ \text{mol}$ of it. The rest, $0.0120\ \text{mol}$ (one for each silver ion), stays as nitric acid: $[\text{H}^+] = 0.0600\ \text{mol L}^{-1}$ and $\text{pH} = 1.22$ .

Watch out. Using all the charge for silver gives $2.013\ \text{g}$ , more silver than the solution holds. Counting every $\text{H}^+$ from the anode gives a pH of $1.03$ .

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