Give the Nernst equation for the $E_{cell}$ of a Nickel-Copper galvanic cell.
(N/A) For a Nickel-Copper galvanic cell,the cell reaction is: $Ni(s) + Cu^{2+}(aq) \rightarrow Ni^{2+}(aq) + Cu(s)$.
According to the Nernst equation,the cell potential $E_{cell}$ is given by:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred).
Thus,the equation is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Ni^{2+}]}{[Cu^{2+}]}$.
According to the Nernst equation,the cell potential $E_{cell}$ is given by:
$E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Ni^{2+}]}{[Cu^{2+}]}$.
Here,$n = 2$ (number of electrons transferred).
Thus,the equation is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log \frac{[Ni^{2+}]}{[Cu^{2+}]}$.
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Calculate the cell potential for the reaction $Mg_{(s)} \mid Mg^{2+}(0.18 \ M) \parallel Ag^{+}(0.01 \ M) \mid Ag_{(s)}$. Given standard electrode potentials are $E^{\circ}_{Mg^{2+}/Mg} = -2.37 \ V$ and $E^{\circ}_{Ag^{+}/Ag} = 0.80 \ V$. (in $V$)
Easy
View SolutionCalculate the cell potential at $298 \ K$ for the following cell:
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)
$Cu_{(s)} | Cu^{2+}(0.1 \ M) || Cu^{2+}(1 \ M) | Cu_{(s)}$
Given: $E_{Cu^{2+}|Cu}^o = 0.34 \ V$ (in $V$)
Medium
View SolutionFor the galvanic cell,
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$
$Zn_{(s)} + Cu^{2+}(0.02 \ M) \rightarrow Zn^{2+}(0.04 \ M) + Cu_{(s)}$
$E_{cell} = ...... \times 10^{-2} \ V \text{ (Nearest integer) }$
$[\text{Use}: E_{Cu^{2+}/Cu}^{0} = 0.34 \ V, E_{Zn^{2+}/Zn}^{0} = -0.76 \ V]$
$[\frac{2.303 \ RT}{F} = 0.059 \ V]$
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