The concentration of Na+ ions in red blood cells is 11 mM and in the surrounding plasma is 140 mM. These values are quite different. Calculate the potential difference, E°', at 37°C (310 K) across the cell membrane because of this concentration gradient assuming the following equilibrium:
Na+(rbc) ⇌ Na+(plasma)
where K = [Na+(plasma)] / [Na+(rbc)] and assume also n = 1 in your calculation.
Group of answer choices:
A. 0.068 V
B. Too small to register on a voltmeter
C. 0.045 V
D. 0.0052 V
E. 0.107 V
Calculate the Ecell value (non-standard state) at 298 K for the cell based on the reaction:
Fe3+(aq) + Cu+(aq) → Fe2+(aq) + Cu2+(aq)
when [Fe3+] = [Cu+] = 1.50 × 10–3 M and [Fe2+] = [Cu2+] = 2.50 × 10–4 M.
Fe3+(aq) + e– → Fe2+(aq) (E° = 0.770 V)
Cu2+(aq) + e– → Cu+(aq) (E° = 0.153 V)
Group of answer choices:
A. 0.617 V
B. 0.507 V
C. 0.923 V
D. 0.709 V
E. 0.599 V
