Question

(a) Compute the voltage at 25˚C of an electrochemical cell consisting of pure cadmium immersed in a 2 × 10^-3 M solution of Cd²⁺ ions and pure iron in a 0.2 M solution of Fe²⁺ ions.

(b)Determine the spontaneous electrochemical reaction for iron (reduced/oxidized) and cadmium (reduced/oxidized).

Answer 1

1) the reaction is given by

Cd²⁺(aq) + 2 e^– --------> Cd(s) E⁰ = -0.403 ( reduction as less negative value )

Fe²⁺(aq) + 2 e^– ---------> Fe(s) E⁰ = -0.44   ( oxidation more -Ve value )

the overall reaction is given by

Cd²⁺(aq) +     Fe(s) ------>    Cd(s) + Fe²⁺(aq)   

E⁰cell = E⁰ reduction - E⁰ oxidation = -0.403 - ( -0.44 ) = 0.037 V

Ecell = E⁰cell - 0.0591/2 log [ Fe²⁺ ] /[Cd²⁺(aq) ]

=  0.037 - 0.02955 log [0.2] / [2 * 10^-3]

= 0.037 - 0.02955 log [100]

=0.037 - 0.02955 *2 = 0.037 - 0.0591 = - 0.0221V

the spontaneous electrochemical reaction for iron (reduced/oxidized) and cadmium (reduced/oxidized).

is given by

Cd²⁺(aq) + 2 e^– --------> Cd(s) reduction

Fe(s) --------->  Fe²⁺(aq) + 2 e^– oxidation