Question
Estimate the cell potential of a Daniel cell having 1.0 M – Zn2+ and originally having 1.0 M – Cu2+ after sufficient ammonia has been added to the cathode compartment to make the NH3 concentration 2.0 M. Given: $E_{Zn^{2 +}|Zn}^{o}$= − 0.76 V, $E_{Cu^{2 +}|Cu}^{o}$= +0.34 V, Kf for Cu(NH3)42+ = 1 × 1012 [2.303 RT/F = 0.06].
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Solution
$Cu^{2 +} + 4NH_{3} \rightleftharpoons Cu\left( NH_{3} \right)_{4}^{2 +},K_{f} = 10^{12}$
1.0 M excess
100% 0 1.0 M
Equ. x 2.0 M 1.0 M
$10^{12} = \frac{1.0}{x \times (2.0)^{4}} \Rightarrow x = \frac{10^{- 12}}{16} = 6.25 \times 10^{- 14}$
Now, cell reaction: Zn + Cu2+ $\rightleftharpoons$Zn2+ + Cu
$E_{cell} = E_{cell}^{o} - \frac{0.06}{2}.\log\frac{\left\lbrack Zn^{2 +} \right\rbrack}{\left\lbrack Cu^{2 +} \right\rbrack} $$$= \lbrack 0.34 - 0.76\rbrack - \frac{0.06}{2}.\log\frac{1.0}{6.25 \times 10^{- 14}} = 0.704\text{ V}$$
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