Question

An electrochemical cell is constructed such that on one side a pure nickel electrode is in contact with a solution containing Ni2+ ions at a concentration of 3 × 10−3 M. The other cell half consists of a pure Fe electrode that is immersed in a solution of Fe2+ ions having a concentration of 0.1 M. At what temperature will the potential between the two electrodes be +0.140 V?

Answer 1

Answer: The temperature at which given potential between the two electrodes is attained is 331.13 K

Explanation:

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

The half reaction follows:

Oxidation half reaction:  $Fe(s)\rightarrow Fe^{2+}(0.1M)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction:  $Ni^{2+}(3\times 10^{-3}M)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.25V$

Net reaction:  $Fe(s)+Ni^{2+}(3\times 10^{-3}M)\rightarrow Fe^{2+}(0.1M)+Ni(s)$

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:

$E^o_{cell}=-0.25-(-0.44)=0.19V$

To calculate the temperature at which the reaction is taking place, we use the Nernst equation, which is:

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

where,

$E_{cell}$ = electrode potential of the cell = +0.140 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.19 V

n = number of electrons exchanged = 2

R = Gas constant = 8.314 J/mol.K

F = Faraday's constant = 96500

T = temperature of the reaction

$[Fe^{2+}]=0.1M$

$[Ni^{2+}]=3\times 10^{-3}M$

Putting values in above equation, we get:

$0.140=0.19-\frac{2.303\times 8.314\times T}{2\times 96500}\times \log(\frac{(0.1)}{(3\times 10^{-3})})\\\\T=331.13K$

Hence, the temperature at which given potential between the two electrodes is attained is 331.13 K