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CHEM 321
Quantitative Analysis
Homework 4 Solutions
1.
An electrochemical cell is assembled from two solutions placed in separate beakers and
buffered to pH 7. One beaker contains amino acids cysteine (0.005 M) and cystine (0.005 M),
the other NAD+ and NADH. The solutions are connected by a salt bridge and furnished with
platinum electrodes. The corresponding half-reactions and formal potentials are as follows:
Cystine + 2H3O+ + 2e¯ → 2 Cysteine + 2H2O
NAD+ + H3O+ + 2e¯ → NADH + H2O
E°' = –0.340 V
E°' = –0.320 V
At what [NAD+]/[NADH] ratio will you measure zero current when shorting the electrodes?
Ecys = –0.340 – (0.0592/2)log [Cred]2/[Cox]
ENAD = –0.320 – (0.0592/2)log [NADH]/[NAD]
[NADH]/[NAD] = x
i = 0 when Ecys = ENAD: –0.340 – (0.0592/2)log (0.0052/0.005) = –0.320 – (0.0592/2)log x
x = 0.0237, [NAD+]/[NADH] = 1/0.0237 = 42.20
2.
Using standard potentials from the textbook, calculate the voltage of each of the following
electrochemical cells:
Fe(s)/FeBr2(0.010 M)//NaBr (0.050 M)/Br2(l)/Pt
Cell Voltage = E(cathode) - E(anode)
E(anode, Fe/Fe2+) = -0.440+(0.05916/2)*log(0.01) = -0.49916 V
E(cathode, Br2, Br-) = 1.065-(0.05916/2)*log(0.05)2 = 1.142 V
E(cell) = 1.142-(-0.49916) = 1.641 V
Cu(s)/Cu(NO3)2(0.020 M)//Fe(NO3)2 (0.050 M)/Fe(s)
Solve as the previous one using standard potentials from the textbook (pp. 808-809)
Hg(l)/Hg2Cl2(s)/KCl(0.060 M)//KCl(0.040 M)/AgCl(s)/Ag(s)
E(anode) = 0.789 + (0.05916/2) log Ksp - 005916 log[Cl-] = 0.789 + (-0.529) - (-0.0723) = 0.332 V
E(cathode) = 0.799 + 0.05916 log(Ksp) - 0.05916 log]Cl-] = 0.799 + (-0.5916) - (-0.0827) = 0.290 V
E(cell) = 0.290 - 0.332 = -0.042 V
3.
The nitrate content in rubble collected at a bomb attack site is determined using a nitrate ionselective electrode. A rubble sample was washed with water to obtain a solution and both it and
the nitrate standards were diluted 20-fold with 0.1 M potassium sulfate before potential
measurements to maintain constant ionic strength. Nitrate standards at 0.0050 M and 0.0100 M
were used, diluted, and gave voltage readings of –0.3496 V and –0.3662 V against S.C.E. (ESCE
= +0.241 V). The sample read –0.3606 against S.C.E. What was the nitrate concentration in
mg/L in the original sample solution before dilution.?
First calculate the slope:
–0.3496 = k – S log(0.005)
–0.3662 = k – S log(0.010)
S = 0.05514
Note that the dilutions do not have to be considered because they were the same for the sample and
internal standards. The 1/20 factor in all concentrations is accommodated in the constant k which we
do not have to calculate because it cancels out when subtracting the equations.
Sample:
–0.3606 = k – 0.05514 log x
–0.3496 = k – 0.05514 log(0.005)
x = 0.0007915 M (to be converted to mg/L in the original wash solution)