(A) The Nernst equation for the cell reaction is given by:$E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$

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(A) The Nernst equation for the cell reaction is given by:$E_{cell} = E^{\circ}_{cell} - \frac{0.059}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$Here,$n = 2$,$E^{\circ}_{cell} = 1.10 \ V$,$[Zn^{2+}] = 0.1 \ M$,and $[Cu^{2+}] = 0.1 \ M$.Substituting these values:$E_{cell} = 1.10 - \frac{0.059}{2} \log \frac{0.1}{0.1}$$E_{cell} = 1.10 - 0.0295 \log(1)$Since $\log(1) = 0$,we get:$E_{cell} = 1.10 - 0 = 1.10 \ V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

For the cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$,the standard $EMF$ value at $25^{\circ}C$ is $1.10 \ V$. If $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,the $EMF$ will be .......... $V$.

A
$1.10$
B
$0.110$
C
$-1.10$
D
$-0.110$

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Electrochemistry

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Nernst equation and ECS

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

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What is the sign of $\Delta G^o$ and the value of $K$ for an electrochemical cell for which $E_{cell}^o = 0.80 \ V$?

Medium

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What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?

MediumMHT CET 2022

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

Medium

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The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

DifficultAIPMT 2007

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For the cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$,the standard $EMF$ value at $25^{\circ}C$ is $1.10 \ V$. If $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,the $EMF$ will be .......... $V$.

Similar Questions

For the redox reaction occurring in a cell: $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$,if $E^o_{cell} = 1.10 \ V$,calculate the value of $E_{cell}$ in $V$. (Given: $2.303 \frac{RT}{F} = 0.0591$)

What is the sign of $\Delta G^o$ and the value of $K$ for an electrochemical cell for which $E_{cell}^o = 0.80 \ V$?

What is the change in potential of the following cell $Zn_{(s)}|Zn^{2+} (1 \ M)||Pb^{2+} (1 \ M)|Pb_{(s)}$ if the concentration of ions at the anode is increased $10$ times?

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

The equilibrium constant of the reaction:$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is:

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The equilibrium constant of the reaction:
$Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is: