(B) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{+2}_{(aq)} + 2Ag_{(s)}$The standard cell potential is: $E^{o}_{cell} = E^{o}_{cat

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$0.46 - \frac{RT}{2F} \ln 10^{7}$

(B) The cell reaction is: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{+2}_{(aq)} + 2Ag_{(s)}$The standard cell potential is: $E^{o}_{cell} = E^{o}_{cathode} - E^{o}_{anode} = 0.80 \ V - 0.34 \ V = 0.46 \ V$Using the Nernst equation: $E_{cell} = E^{o}_{cell} - \frac{RT}{nF} \ln Q$Here,$n = 2$ and $Q = \frac{[Cu^{+2}]}{[Ag^{+}]^2} = \frac{10^{-3}}{(10^{-5})^2} = \frac{10^{-3}}{10^{-10}} = 10^{7}$Substituting the values: $E_{cell} = 0.46 - \frac{RT}{2F} \ln 10^{7}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be

A
$0.46 \ V$
B
$0.46 - \frac{RT}{2F} \ln 10^{7}$
C
$0.46 + \frac{RT}{2F} \ln 10^{7}$
D
$0.46 - \frac{RT}{2F} \ln 10^{2}$

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Nernst equation and ECS

For the reaction; $2A_{(s)} + B_{(aq)}^{2+} \to 2A_{(aq)}^{+} + B_{(s)}$,the Nernst equation for the $EMF$ of the cell is:

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What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)

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The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is

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For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

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The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

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$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$$E_{cell}$ will be

Similar Questions

For the reaction; $2A_{(s)} + B_{(aq)}^{2+} \to 2A_{(aq)}^{+} + B_{(s)}$,the Nernst equation for the $EMF$ of the cell is:

What is the potential of a cell containing two hydrogen electrodes,the negative one in contact with $10^{-8} \ M \ H^{+}$ and the positive one in contact with $0.025 \ M \ H^{+}$? (in $V$)

The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is

For the cell $Zn | Zn^{2+}_{(aq)} || Cu^{2+}_{(aq)} | Cu$,the standard cell potential $E^o$ is $1.10 \ V$ at $25^o \ C$. What is the order of magnitude of the equilibrium constant $K$ for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$?

The $emf$ of the cell $Tl_{(s)} | Tl^{+}_{(aq)} (0.0001 \ M) || Cu^{2+}_{(aq)} (0.01 \ M) | Cu_{(s)}$ is $0.83 \ V$. How can the $emf$ of this cell be increased?

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$Cu_{(s)} | Cu^{+2}(aq, 10^{-3} M) || Ag^{+}(aq, 10^{-5} M) | Ag_{(s)}$
If $E^{o}_{Cu^{+2}/Cu} = +0.34 \ V$
$E^{o}_{Ag^{+}/Ag} = +0.80 \ V$
$E_{cell}$ will be