ElectrochemistryHard
Question
Zinc granules are added in excess to a 500 ml of 1.0 M nickel nitrate solution at 25°C until the equilibrium is reached. If the standard reduction potential of Zn2+|Zn and Ni2+|Ni are − 0.75 V and −0.24 V, respectively, the concentration of Ni2+ in solution at equilibrium is (2.303 RT/F = 0.06)
Options
A.1.0 × 10−17 M
B.1.0 × 1017 M
C.5 × 10−17 M
D.2 × 10−17 M
Solution
$Zn + Ni^{2 +} \rightleftharpoons Zn^{2 +} + Ni$
$E_{cell}^{o} = \frac{0.06}{2}.\log K_{eq}$
⇒ (–0.24) – (–0.75) = 0.03 log Keq
∴ Keq = 1017 ⇒ It means that Ni2+ will react almost completely and [Zn2+] ≈ 1.0 M
Now, $10^{17} = \frac{1.0}{\left\lbrack Ni^{2 +} \right\rbrack} \Rightarrow \left\lbrack Ni^{2 +} \right\rbrack = 10^{- 17}\text{ M}$
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