Question

If the concentration of Ag+ is 0.0115 M, the concentration of H+ is 0.355 M, and the pressure of H2 is 1.00 atm, calculate the cell potential at 25.0°C. The standard reduction potentials are: $$ Eo = 0.80 V $$ Eo = 0.00 V

Answer 1

Answer: The cell potential is 0.712 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, silver will undergo reduction reaction will get reduced. Hydrogen will undergo oxidation reaction and will get oxidized

Half reactions for the given cell follows:

Oxidation half reaction: $H_2(1.00atm)\rightarrow 2H^{+}(0.355M)+2e^-;E^o_{H^{+}/H_2}=0.00V$

Reduction half reaction: $Ag^{+}(0.0115M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.80V$   ( × 2 )

Net reaction: $H_2(1.00atm)+2Ag^{+}(0.0115M)\rightarrow 2H^{+}(0.355M)+2Ag(s)$

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:

$E^o_{cell}=0.80-(0.00)=0.80V$

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

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

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.80 V

n = number of electrons exchanged = 2

$[Ag^{+}]=0.0115M$

$[H^{+}]=0.355M$

Putting values in above equation, we get:

$E_{cell}=0.80-\frac{0.059}{2}\times \log(\frac{(0.355)^2}{(0.0115)^2})\\\\E_{cell}=0.712V$

Hence, the cell potential is 0.712 V