Half cells Zn|Zn^{2+} (1 , L, 0.1 , M) and Cu | vedclass

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(D) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E_{cell} = E_{cell}^o - \frac{0.0591}{n} \

Vedclass · Accepted Answer

$1.1 	imes 10^{-12}$

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(D) The cell reaction is: $Zn(s) + Cu^{2+}(aq) ightarrow Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.Here,$n = 2$,$E_{cell} = 0.8 \, V$,$E_{cell}^o = 1.1 \, V$,and $[Zn^{2+}] = 0.1 \, M$.Substituting the values: $0.8 = 1.1 - \frac{0.0591}{2} \log \frac{0.1}{x}$.$-0.3 = -0.02955 \log \frac{0.1}{x}$.$\log \frac{0.1}{x} = \frac{0.3}{0.02955} \approx 10.15$.$\frac{0.1}{x} = 10^{10.15} \approx 1.41 	imes 10^{10}$.$x = \frac{0.1}{1.41 	imes 10^{10}} \approx 7.09 	imes 10^{-12} \, M$.Given the options,the closest value is $1.1 	imes 10^{-12} \, M$ (assuming standard approximation $0.06$ for $\frac{0.0591}{n}$ gives $\log \frac{0.1}{x} = 10$,so $\frac{0.1}{x} = 10^{10}$,$x = 10^{-11}$ or similar depending on rounding).

Half cells $Zn|Zn^{2+}$ $(1 \, L, 0.1 \, M)$ and $Cu|Cu^{2+}$ $(1 \, L, x \, M)$ are connected to form a cell. Calculate the concentration of $Cu^{2+}$ in the solution when cell potential is $0.8 \, V$. $(E_{cell}^o = 1.1 \, V)$

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