Question

Calculate the theoretical potential of the following cell. Indicate whether the reaction will proceed spontaneously in the direction considered (oxidation on the left, reduction on the right) or whether an external voltage source is needed to force this reaction to occur.

Pt, H2(757 torr)|HCl(2.00×10-4 M) parallel to Ni2+(0.0400 M)|Ni

the answer is

-0.072 V; External voltage needed

can anyone explain to me how i get this answer ?

Answer 1

Answer: The theoretical potential of the given cell is -0.072 V

Explanation:

The given chemical cell follows:

$Pt(s)|H_2(g,757torr)|HCl(aq,2.00\times 10^{-4}M)||Ni^{2+}(aq,0.0400M)|Ni(s)$

Oxidation half reaction: $H_2(g,757torr)\rightarrow 2H^{+}(aq,2.00\times 10^{-4}M)+2e^-;E^o_{2H^{+}/H_2}=0.0V$

Reduction half reaction: $Ni^{2+}(aq,0.0400M)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.25V$

Net cell reaction: $H_2(g,757torr)+Ni^{2+}(aq,0.0400M)\rightarrow 2H^{+}(aq,2.00\times 10^{-4}M)+Ni(s)$

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:

$E^o_{cell}=-0.25-(0.0)=-0.25V$

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

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[H^{+}]^2}{[Ni^{2+}]\times p_{H_2}}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = -0.25 V

n = number of electrons exchanged = 2

$[H^{+}]=2.00\times 10^{-4}M$

$[Ni^{2+}]=0.0400M$

$p_{H_2}=757torr=0.996atm$       (Conversion factor:  1 atm = 760 torr)

Putting values in above equation, we get:

$E_{cell}=-0.25-\frac{0.059}{2}\times \log(\frac{(2.00\times 10^{-4})^2}{0.0400\times 0.996})\\\\E_{cell}=-0.072V$

Hence, the theoretical potential of the given cell is -0.072 V