Suppose you construct the following galvanic cell:
Fe(s)|Fe2+(aq)||NAD+(aq)|NADH(aq)
Fe2+ + 2e- → Fe Eo = -0.44V
NAD+ + 2e- + 2H+ → NADH +
H+ Eo' = -0.320V
(note the different standard states)
A) Will the cell generate current at biochemical standard state?
At chemical standard state? At T = 4 oC, pH = 7? At T =
97 oC, pH = 7? Justify your answers.
B) How many protons can be moved across a membrane from pH 7.5
to...
Consider the following cell at 280 K:
Fe | Fe2+ (0.573) || Cd2+ (1.063) | Cd
which has a standard cell potential of 0.0400 V. What will be
the potential of the cell be when [Fe2+] changes by
0.341 M?
Consider the following cell at 277 K:
Fe | Fe2+ (0.567) || Cd2+ (1.063) | Cd
which has a standard cell potential of 0.0400 V. What will be
the potential of the cell be when [Fe2+] changes by
0.305 M?
The answer is 0.0383 V
Consider the following half reactions at 298 K
Fe2+ + 2 e- →
Fe Eo = -0.441 V
Cd2+ + 2 e- →
Cd Eo = -0.403 V
A galvanice cell based on these half reactions is set up under
standard conditions where each solution is 1.00 L and each
electrode weighs exactly 100.0 g. How much will the Cd electrode
weigh when the nonstandard potential of the cell is 0.02804 V?
Consider the following half reactions at 298 K
Fe2+ + 2 e- →
Fe Eo = -0.441 V
Cd2+ + 2 e- →
Cd Eo = -0.403 V
A galvanice cell based on these half reactions is set up under
standard conditions where each solution is 1.00 L and each
electrode weighs exactly 100.0 g. How much will the Cd electrode
weigh when the nonstandard potential of the cell is 0.02880 V?
The answer is 139g.
During the corrosion of iron, iron is oxidized (Fe → Fe2+ + 2e
−) and molecular oxygen is reduced (O2 + 2H2O + 4e −→ 4HO−). The E
◦ for the reduction of Fe2+ to Fe is −0.44 V, the E ◦ for the
reduction of O2 is 0.40 V, and the E ◦ for the reduction of Au3+ to
Au (Au3+ + 3e −→ Au) is 1.50 V. What would E ◦ cell be for the
oxidation of gold...
Consider an electrochemical cell based on the following overall
reaction, Fe(s) + 2Ag+(aq) Fe2+(aq) + 2Ag(s) Fe2+(aq) + 2e–
Fe(s), ℰ° = –0.44 V Ag+(aq) + e– Ag, ℰ° = 0.80 V Calculate the cell
potential (in V) for this reaction at 25oC when the concentration
of Ag+ ions is 0.050 M and the concentration of Fe2+ ions is 1.50
M. a. +1.32 V b. +1.50 V c. +1.20 V d. -1.32 V e. -1.20 V