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} \

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$1.1 	imes 10^{-12}$

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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 $0.8 \, V$. $(E_{cell}^o = 1.1 \, V)$

Similar Questions

Write the Nernst equation for the $E_{cell}$ of a Daniell cell.

The $emf$ of the following three galvanic cells:
$1. Zn|Zn^{2+} (1 \ M)||Cu^{2+} (1 \ M)|Cu$
$2. Zn|Zn^{2+} (0.1 \ M)||Cu^{2+} (1 \ M)|Cu$
$3. Zn|Zn^{2+} (1 \ M)||Cu^{2+} (0.1 \ M)|Cu$
are represented by $E_1, E_2, E_3$. Which of the following statements is true?

If $E^{\circ}(Al_{(aq)}^{+3} \mid Al_{(s)}) = -1.66 \ V$. What is the potential of $Al_{(s)} \rightarrow Al_{(aq)}^{+3}(0.1 \ M) + 3e^-$ at $298 \ K$?

Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

For the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$

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