Question

1) When the Ag⁺ concentration is 1.08 M, the observed cell potential at 298K for an electrochemical cell with the following reaction is 1.654V. What is the Zn²⁺ concentration?

2Ag⁺(aq) + Zn(s) > 2Ag(s) + Zn²⁺(aq)

Answer: ___ M

2) A concentration cell similar to the one shown is composed of two Cr electrodes and solutions of different Cr³⁺concentrations. The left compartment contains 1.32 M Cr³⁺, and the right compartment contains 0.250 M Cr³⁺.


Calculate the cell potential for this reaction at 298 K.

____ volts

In this chromium concentration cell, the compartment on the left is the __anode or cathode__, and the compartment on the right is the _anode or cathode__.

Answer 1

Ans:

Data collected	Standard reduction potential = Eo at 298 K
Ag+ + e --> Ag	0.80 V
Zn2+ + 2e---> Zn	-0.76 V
Cr3+ + 3e --> Cr	-0.74 V

Ans 1:

Cell reaction: Zn(s) + 2Ag⁺(aq) --> Zn²⁺(aq) + 2Ag(s)

No. of electrons transferred, n = 2

Q = Reaction quotient = [Zn²⁺(aq)]/[Ag⁺(aq)]² = [Zn²⁺(aq)]/(1.08)²

Cell potential is given by Nernst equation;

E_cell = E^o_cell – (2.303RT/nF) log Q

Where,

E_cell = cell voltage = 1.654 V

E^o_cell = cell voltage under standard conditions = E^ocathode-E^oanode = 0.80-(-0.76) = 1.56V

R = Universal gas constant = 8.314 J/(K.mol)

T = Temperature in kelvin = 298 K

F = Faraday constant = 96500 C/mol

E_cell = E^o_cell – (2.303RT/nF) log Q

1.654 = 1.56 – {(2.303x8.314x298)/(2x96500)}log {[Zn²⁺(aq)]/(1.08)²}

[Zn²⁺(aq)] = 7.59 x 10^-4 M

Ans 2:

Oxidation half-cell (anode) = Cr(s) ---> Cr³⁺(aq, 0.250M) + 3e

Reduction Half-cell (cathode) = Cr³⁺(aq, 1.32M) + 3e --> Cr(s)

Cell reaction: Cr(s) + Cr³⁺(aq, 1.32M) ---> Cr³⁺(aq, 0.250M) + Cr(s)

Reaction quotient, Q = [Cr³⁺(aq)]/[ Cr³⁺(aq, 1.32M)] = 0.250/1.32 = 0.1894

No. of electrons exchanged, n = 3

Cell potential is given by Nernst equation;

E_cell = E^o_cell – (2.303RT/nF) log Q

Where,

E_cell = cell voltage = ?

E^o_cell = cell voltage under standard conditions = 0 for concentration cell

R = Universal gas constant = 8.314 J/(K.mol)

T = Temperature in kelvin = 298 K

F = Faraday constant = 96500 C/mol

E_cell = E^o_cell – (2.303RT/nF) log Q

E_cell = 0 – {(2.303x8.314x298)/(3 x 96500)}log(0.1894) = 0.0142 V

Left compartment = 1.32 M = Cathode

Right compartment = 0.250 M = Anode