Calculate the cell potential at $298 \ K$ for the following cell:
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$
$Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cu^{2+} (0.3 \ M) | Cu_{(s)} \quad [E_{cell}^{o} = 1.1 \ V]$
The cell reaction is: $Zn_{(s)} + Cu^{2+} (aq) \rightarrow Zn^{2+} (aq) + Cu_{(s)}$
Number of electrons transferred,$n = 2$.
Using the Nernst equation at $298 \ K$:
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Substitute the given values:
$E_{cell} = 1.1 - \frac{0.0591}{2} \log \frac{0.6}{0.3}$
$E_{cell} = 1.1 - 0.02955 \log 2$
Since $\log 2 \approx 0.3010$:
$E_{cell} = 1.1 - (0.02955 \times 0.3010)$
$E_{cell} = 1.1 - 0.00889$
$E_{cell} \approx 1.0911 \ V$
Number of electrons transferred,$n = 2$.
Using the Nernst equation at $298 \ K$:
$E_{cell} = E_{cell}^{o} - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$
Substitute the given values:
$E_{cell} = 1.1 - \frac{0.0591}{2} \log \frac{0.6}{0.3}$
$E_{cell} = 1.1 - 0.02955 \log 2$
Since $\log 2 \approx 0.3010$:
$E_{cell} = 1.1 - (0.02955 \times 0.3010)$
$E_{cell} = 1.1 - 0.00889$
$E_{cell} \approx 1.0911 \ V$
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Similar Questions
Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$ $C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$
$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.
| $A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ | $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$ |
| $C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ | $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$ |
$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.
In which of the following Galvanic cells is the $emf$ maximum? (Given: $E_{Mg^{2+} \mid Mg}^0 = -2.36 \ V$ and $E_{Cl_2 \mid 2 Cl^{-}}^0 = +1.36 \ V$)
Calculate $E_{cell}$ of the reaction $Mg_{(s)} + 2Ag^{+}_{(aq)} (0.0001 \ M) \to Mg^{2+}_{(aq)} (0.100 \ M) + 2Ag_{(s)}$ in $V$. If $E^o_{cell} = 3.17 \ V$.
Medium
View SolutionWhich of the following will increase the voltage of the cell represented by the equation
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$
$Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$
Medium
View SolutionGiven the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]
DifficultJEE MAIN 2019
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