Question

onsider the galvanic cell described by (N and M are metals): N(s)|N2+(aq)||M+(aq)|M(s) If Eocathode = 0.576 V and Eoanode = 1.129 V, and [N2+(aq)] = 0.307 M and [M+(aq)] = 0.973 M, what is Ecell, using the Nernst equation? ____ V

Answer 1

Answer: The cell potential of the given cell is -0.567 V

Explanation:

The given chemical cell follows:

$N(s)|N^{2+}(aq.)||M^{+}(aq.)|M(s)$

Oxidation half reaction:  $N(s)\rightarrow N^{2+}(aq.)+2e^-;E^o_{N^{2+}/N}=1.129V$

Reduction half reaction:  $M^{+}(aq.)+2^-\rightarrow M(s);E^o_{M^+/M}=0.576V$     ( × 2)

Net cell reaction: $N(s)+2M^{+}(aq.)\rightarrow N^{2+}(aq.)+2M(s)$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.576-(1.129)=-0.553V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[N^{2+}]}{[M^{+}]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ? V

$E^o_{cell}$ = standard electrode potential of the cell = -0.553 V

n = number of electrons exchanged = 2

$[M^{+}]=0.973M$

$[N^{2+}]=0.307M$

Putting values in above equation, we get:

$E_{cell}=-0.553-\frac{0.059}{2}\times \log(\frac{0.307}{(0.973)^2})\\\\E_{cell}=-0.567V$

Hence, the cell potential of the given cell is -0.567 V