Question

A galvanic cell is based on the following half-reactions at 279 K: Ag+ + e- → Ag Eo = 0.803 V H2O2 (aq) + 2 H+ + 2 e- → 2 H2O Eo = 1.78 V What will the potential of this cell be when [Ag+] = 0.559 M, [H+] = 0.00393 M, and [H2O2] = 0.863 M?

Answer 1

Answer: The potential of the given cell is 0.856 V

Explanation:

The substance having highest positive $E^o$ potential will always get reduced.

Half reactions for the given cell follows:

Oxidation half reaction: $Ag\rightarrow Ag^{+}+e^-;E^o_{Ag^{+}/Ag}=0.803V$     ( × 2)

Reduction half reaction: $H_2O_2(aq.)+2H^{+}+2e^-\rightarrow 2H_2O;E^o=1.78V$

Net reaction:  $2Ag(s)+H_2O_2(aq.)+2H^{+}\rightarrow 2Ag^{+}+2H_2O$

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:

$E^o_{cell}=1.78-(0.803)=0.977V$

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

$E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Ag^{+}]^2}{[H^{+}]^2[H_2O_2]}$

where,

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

$E^o_{cell}$ = standard electrode potential of the cell = +0.977 V

R = Gas constant = 8.314 J/K mol

T = temperature = 279 K

F = Faraday's constant = 96500 C

n = number of electrons exchanged = 2

$[Ag^{+}]=0.559M$

$[H^{+}]=0.00393M$

$[H_2O_2]=0.863M$

Putting values in above equation, we get:

$E_{cell}=0.977-\frac{2.303\times 8.314\times 279}{2\times 96500}\times \log(\frac{(0.559)^2}{(0.00393)^2\times (0.863)})\\\\E_{cell}=0.856V$

Hence, the potential of the given cell is 0.856 V