Question

Using activities, calculate the concentration Cu2+ if the overall potential of the cell is 0.572V. Show all the work and state your final result with 3 sig fig. The standard reduction potentials are: E Cu2+/Cu(s)=0.339V and ENi2+/Ni(s)=-0.236V. Assume an activity coefficient for Cu2+=0.690 and for Ni2+=0.0675

Answer 1

In this problem a cell is made up of Cu and Ni electrodes,

Standard reduction potential: Cu²⁺ / Cu = 0.339V, and

Ni²⁺/ Ni = -0.236 V , it has comparatively lower reduction potential, so it will act like an anode and have oxidation on it, with standard oxidation potential = +0.236 V, while the other electrode is acting as cathod.

The reaction involving in this problem is:

At Anode- $Ni(s) \rightarrow Ni^{^+2} + 2e^{-}$

At cathod - $Cu^{^+2} + 2e^{-} \rightarrow Cu(s)$

Overall reaction - $Ni(s) + Cu^{^+2} \rightarrow Ni^{^+2} + Cu(s)$

Overall E^o_Cell=(0.236 + 0.339) V = 0.575 V

Formula of EMF or  overall potential of the cell:

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

Since we have not provided with temperature of the reaction, so we are assuming reaction is taking place at room temperature, for that, $\frac{2.303RT}{nF} = \frac{0.059}{n}$

therefore, $E_{cell} = E^{o}_{cell} - \frac{0.059}{n} log \frac{(Ni^{2+})_{a}}{(Cu^{2+})_{a}}$

here, n= number of electrons involving in this process = 2 & E_cell = 0.527 V

$0.527 = 0.575 - \frac{0.059}{2} log \frac{(Ni^{2+})_{a}}{(Cu^{2+})_{a}}$

$log \frac{(Ni^{2+})_{a}}{(Cu^{2+})_{a}} = 1.627$

$\frac{(Ni^{2+})_{a}}{(Cu^{2+})_{a}} = 42.364$

(Ni²⁺)_a and (Cu²⁺)_a are the activities of Ni²⁺ and Cu²⁺ ions, respectively, and

$Activitity = Concentration \times Activity coefficient$

activity coefficient for Cu²⁺=0.690 and for Ni²⁺=0.0675.

$\frac{ [Ni^{2+}] \times 0.0675}{ [Cu^{2+}] \times 0.690} =42.364$

$\frac{ [Ni^{2+}] }{ [Cu^{2+}] } =433.054$

$[Cu^{2+}] =0.0023 \times [Ni^{2+}]$