Use a K+ NO3- salt bridge (100 mL on both sides)

Question

Use a K+ NO3- salt bridge (100 mL on both sides)

1M LNO3 1M R(NO3)2

(d) If 1.0 gram of solid NaCl is added to each half-cell, the cell voltage increases. Explain why. 

(e) If 20.0 mL of distilled water is added to both half-cells, the cell voltage increases. Explain why. 

(f) If a current of 11.4 amps is passed through a solution of LNO₃ for 8.1 hours, how many grams of metal L (MM = 67.1) are produced?

Answer 1

(f) 231 g of L. Parts (d) and (e) are Nernst-equation reasoning, and the two observations together actually tell you which way the cell runs — that is the nice part of this problem.

First, deduce the cell reaction from part (e)

E = E° − (0.0592/n) log Q. Voltage rises when Q falls.

Adding water dilutes both half-cells by the same factor f (here 100/120 = 0.833). Watch what that does to Q depending on direction:

If the reaction were R(s) + 2L⁺ → R²⁺ + 2L(s), then Q = [R²⁺]/[L⁺]² and dilution gives Q_new = Q/f — Q increases, so voltage would drop.

If instead 2L(s) + R²⁺ → 2L⁺ + R(s), then Q = [L⁺]²/[R²⁺] and dilution gives Q_new = f·Q — Q decreases, so voltage rises.

You were told the voltage rises, so the second direction is the correct one: L is oxidised at the anode, R²⁺ is reduced at the cathode.

(e) Why dilution raises the voltage

The cell reaction produces two moles of dissolved ions (2 L⁺) for every one it consumes (R²⁺). Diluting shifts an equilibrium toward the side with more dissolved particles, so it favours the forward reaction and pushes E up.

Quantitatively: Q goes from 1.00 to 0.833, log Q = −0.079, and E rises by (0.0296)(0.079) = about 2 mV. A real increase, but a small one — which is what you should expect from a modest dilution.

(d) Why adding NaCl raises the voltage

The Na⁺ is a spectator; the Cl⁻ is what matters. Chloride precipitates L⁺ as insoluble LCl, the way Cl⁻ does with Ag⁺, Pb²⁺, and Hg₂²⁺.

That removes L⁺ from solution — and L⁺ is a product of the cell reaction. Removing a product lowers Q and raises E, exactly as Le Châtelier predicts: the cell responds by driving the forward reaction harder.

(If chloride had instead precipitated R²⁺, a reactant, the voltage would have fallen. The direction of the observed change is what tells you which ion the chloride grabbed.)

(f) Electrolysis calculation

Charge passed: Q = It = (11.4 A)(8.1 h)(3600 s/h) = 3.324 × 10⁵ C

Moles of electrons: 3.324 × 10⁵ / 96 485 C/mol = 3.445 mol e⁻

From the formula LNO₃, nitrate is 1−, so L is L⁺ and the reduction is L⁺ + e⁻ → L — one electron per atom.

m = (3.445 mol)(67.1 g/mol) = 231 g of L

The step people miss is the charge on L. Reading it off the nitrate formula is the reliable way — if L had been L²⁺, the answer would be half this.