Question

A galvanic cell (battery) is represented by the shorthand below. Look up the standard half cell potentials in your textbook or on the internet. (Make sure you select the half reactions that have all the species required, and no other species.)

            Cd(s) / Cd²⁺ // IO₃^- , H+, Cl^-, ICl₂^- / Pt(s)

(a) Calculate E^o for this battery.

(b) What is the oxidizing agent (chemical formula of the species, not just the element)?

What is the reducing agent (chemical formula of the species, not just the element)?

            What species is/are found in the cathode compartment?

(c) Determine the overall balanced chemical equation for the battery reaction.

(d) Use Nernst eqn and the concentrations below, calculate expected cell potential.

[Cd²⁺] = 0.0250 M     [IO₃^-] = 0.100     pH = 4.00      [Cl^-] = 0.500 M        [ICl₂^-] = 1.75 x 10^-5 M

Answer 1

anode reaction : oxidation

Cd ---> Cd+2 + 2e-

cathode reaction : reduction

I03- + 2Cl- + 6H+ + 4e- --> ICl2- + 3H20

so

overall balanced chemical equation is

2 Cd(s) + I03- + 2 Cl- + 6H+ ---> 2 Cd+2   + ICl2- + 3H20

reducing agent is which undergoes ozidation

so

Cd is the reducing agent

oxidzing agent is the one which undergoes reduction

so

I03- is the oxidizing agent

now

Eo cell = Eo cathode - Eo anode

Eo cell = 1.24 - (-0.4)

Eo cell = 1.64

so

Eo for this battery is 1.64 V

now

2 Cd(s) + I03- + 2 Cl- + 6H+ ---> 2 Cd+2   + ICl2- + 3H20

the reaction quotient is given by

Q = [Cd+2]^2 [ICl2-] / [ I03-] [Cl-]^2 [ H+]^6

now

[H+] = 10^-pH = 10-4

now

Q = [ 0.025]^2 [1.75 x 10-5 ] / [0.1] [0.5]^2 [10-4]^6

Q = 4.375 x 10^17

now

according to nernst equation

E = Eo - ( 0.0592 / n) log Q

here

n = 4 as four electrons are transferred

so

E = 1.64 - ( 0.0592 / 4) log 4.375 x 10^17

E = 1.379

so

the expected cell potential is 1.379 V