Different concentrations of ions on either side of a cell membrane constitute a concentration cell, which is described by the Nernst equation given below. What is the electrochemical potential across a membrane when the sodium ion concentration is 0.10 M on one side of the membrane and 0.010 M on the other side, and the temperature is 37°C (body temperature) ?
(Given: Na⁺ + e^- $\rarr$ Na, $E _ { \mathrm { red } } ^ { 0 }$ = -2.71 V; F = 96,485 C; 1 J = 1 C V; R = 8.315 J / (mol K) )
$E = E { } ^ { \circ } - \frac { R T } { n F } \ln Q$

Question

Different concentrations of ions on either side of a cell membrane constitute a concentration cell, which is described by the Nernst equation given below. What is the electrochemical potential across a membrane when the sodium ion concentration is 0.10 M on one side of the membrane and 0.010 M on the other side, and the temperature is 37°C (body temperature) ?
(Given: Na⁺ + e^-

\rarr

Na,

E _ { \mathrm { red } } ^ { 0 }

= -2.71 V; F = 96,485 C; 1 J = 1 C V; R = 8.315 J / (mol K) )

E = E { } ^ { \circ } - \frac { R T } { n F } \ln Q

Quizplus · Accepted Answer

The electrochemical potential across a membrane can be calculated using the Nernst equation:$E = E { } ^ { \circ } - \frac { R T } { n F } \ln Q$where:$E$ is the electrochemical potential in volts$E { } ^ { \circ }$ is the standard reduction potential in volts$R$ is the ideal gas constant in J/(mol K)$T$ is the temperature in Kelvin$n$ is the number of electrons transferred in the reaction$F$ is the Faraday constant in C/mol$Q$ is the reaction quotientIn this case, the reaction is:$Na { } ^ { + } + e { } ^ { - } \rightarrow Na$and the standard reduction potential is -2.71 V. The temperature is 37°C, which is 310 K. The number of electrons transferred is 1. The Faraday constant is 96,485 C/mol. The reaction quotient is:$Q = \frac { [Na { } ^ { + } ] } { [Na] }$where:$[Na { } ^ { + } ]$ is the concentration of sodium ions on the side of the membrane with the higher concentration$[Na]$ is the concentration of sodium ions on the side of the membrane with the lower concentrationIn this case, $[Na { } ^ { + } ]$ is 0.10 M and $[Na]$ is 0.010 M. Therefore, the reaction quotient is:$Q = \frac { 0.10 M } { 0.010 M } = 10$Substituting these values into the Nernst equation, we get:$E = -2.71 V - \frac { 8.315 J / (mol K) \times 310 K } { 1 \times 96,485 C/mol } \ln 10$$E = -2.71 V - 0.0615 V \ln 10$$E = -2.71 V - 0.0615 V \times 2.303$$E = -2.71 V - 0.141 V$$E = -2.85 V$Therefore, the electrochemical potential across the membrane is -2.85 V.

Different Concentrations of Ions on Either Side of a Cell →\rarr→

Different Concentrations of Ions on Either Side of a Cell $\rarr$