ASAP J.R. potential table is given below.

Question

Answer the following questions regarding the electrochemical cell shown. A portion of the standard reduction potential table is given below.

L⁺ + e^- → L     E° = 0.20 V

R²⁺ + 2e^- → R   E° = 0.90 V

Use a K+ and NO3- saltbridge

(k+) LEFT: 100 mL ; 1 M LNO3

(NO3-) RIGHT: 100 mL ; 1 M 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 cells, will the cell voltage increase, decrease, remain the same? Explain.

(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

Set the cell up first, because (d) and (e) both hang on it.

R2+ + 2e- -> R has the more positive standard potential (0.90 V), so R2+ is what gets reduced: R is the cathode. That leaves L+ + e- -> L (0.20 V) to run in reverse, so L is the anode and L metal is oxidized.

Overall: 2 L(s) + R2+(aq) -> 2 L+(aq) + R(s)
E(cell, standard) = 0.90 - 0.20 = 0.70 V

Nernst at 25 C, with n = 2:
E = 0.70 - (0.0592/2) log Q, where Q = [L+]^2 / [R2+]

Keep that Q in front of you. Parts (d) and (e) are both just "what happens to Q."

(d) 1.0 g NaCl into each half-cell

Sodium ion and nitrate are spectators, so chloride is the only thing that can do anything. The fact that the voltage rose is your evidence: chloride precipitated L+ as insoluble LCl, pulling L+ out of the anode solution. (Insoluble chlorides belong to the classic 1+ cations, so a 1+ L fits; RCl2 is soluble, so the cathode side is untouched.)

L+ is a product, so it lives in the numerator of Q. Lower [L+] gives a smaller Q, log Q goes negative, you subtract a negative number, and E goes up. That matches the observation.

Notice the direction of the logic: the problem hands you the result and asks you to infer the chemistry. If both chlorides had been soluble, adding NaCl to both sides would have changed essentially nothing.

(e) 20.0 mL of distilled water into both cells

The tempting answer is "no change, both sides were diluted equally." That is the trap.

Both go from 100 mL to 120 mL, so both concentrations are multiplied by 100/120 = 0.833. But L+ is squared in Q and R2+ is not, so the two dilutions do not cancel:

Q = (0.833)^2 / (0.833) = 0.833

Q dropped from 1.00 to 0.833, so
E = 0.70 - (0.0296) log(0.833) = 0.70 - (0.0296)(-0.0792) = 0.702 V

The voltage increases, by about 2 mV. Tiny, but the sign is the whole point: equal dilution only cancels when the exponents in Q match, and here they are 2 and 1. Any time the two ions carry different charges, check this instead of assuming symmetry.

(f) 11.4 A through LNO3 for 8.1 hours

Charge: q = I t = (11.4 A)(8.1 h)(3600 s/h) = 332,424 C

Electrons: 332,424 / 96,485 = 3.445 mol e-

L+ + e- -> L is one electron per atom, so 3.445 mol of L. This is where the 1+ charge earns its keep; if L were 2+ you would divide by 2 here.

Mass: (3.445 mol)(67.1 g/mol) = 231 g of L

One habit worth taking from this problem: write Q out with its exponents before you reason about any concentration change. Part (e) gets answered wrong from intuition almost every time, and the exponents are the only thing that gives it away.