Question

1. A galvanic cell is based on the following half-reactions at 285 K:

Ag⁺ + e^- → Ag     E^o = 0.803 V  
H₂O₂ (aq) + 2 H⁺ + 2 e^- → 2 H₂O     E^o = 1.78 V

What will the potential of this cell be when [Ag⁺] = 0.571 M, [H⁺] = 0.00341 M, and [H₂O₂] = 0.895 M?

Please show full work .

Answer 1

The given reduction reactions are

Ag+ (aq) + e- -------> Ag (s)...............E^o = 0.803 V

H2O2 (aq) + 2H+ (aq) + 2e- ----------> 2 H2O (l) ...........E^o = 1.78 V

Since reduction potential of reaction involving H2O2 is higher, It would occur at cathode and Ag electrode would behave as an anode.

Therefore we have

Reaction at cathode : H2O2 (aq) + 2H+ (aq) + 2e- ----------> 2 H2O (l) .........E^o = 1.78 V

Reaction at anode : Ag (s) --------> Ag+ (aq) + e- ..............................E^o = - 0.803 V ( Since the reaction is reversed, sign of E^o would also change)

Let's find the standard emf, E^o cell for the reaction.

E^o cell = E^o ​cathode - E^o ​anode

E^o ​cell = 1.78 V - ( 0.803 V)

E^o ​cell = 0.977 V

The potential of the cell can be found out using Nernst equation.

The overall cell reaction is

H2O2 (aq) + 2H+ (aq) + 2Ag(s) ------------> 2 H2O (l) + 2 Ag+ (aq) ...........( we multiply anode reaction by 2 to make the number of electrons equal and then cancel out the electrons to get the overall reaction)

Let's apply Nernst equation

Ecell = Eo cell - RT/nF * ln[ Q]

Where, R is gas constant , 8.314 J/ mol K

n is number of electrons transferred which is 2

F is Faraday's constant , 96500

T is temperature in Kelvin 285 K

Q is reaction quotient which can be written as Q = [Ag+]^2 / [H2O2] [ H+]^2

Let's substitute above values in Nernst equation.

Ecell = 0.977 V - (8.314 J/ mol K * 285 K) / 2 * 96500 * ln [ (0.571)^2 / (0.895) ( 0.00341)^2 ]

E cell = 0.977 V - 0.012277 * ln ( 31328.56)

E cell = 0.977 V - ( 0.012277 * 10.35)
E cell = 0.977 V - 0.127 V

E cell = 0.850 V

The potential of this cell is 0.850 V