(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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Class 12

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Electrochemistry

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Nernst equation and ECS

For the cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

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Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

MediumKVPY 2009

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$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

Medium

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Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

Medium

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In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

MediumAP EAMCET 2025

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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 cell $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$ $(E^{\circ}_{cell} = 1.10 \ V)$ at $298 \ K$,when the cell is completely discharged,what is the ratio of concentrations $\frac{[Zn^{2+}]}{[Cu^{2+}]}$?

Consider the following electrochemical cell,$Zn_{(s)} + 2Ag^{+}(0.04\, M) \longrightarrow Zn^{2+}(0.28\, M) + 2Ag_{(s)}$. If $E_{\text{cell}}^{\circ} = 2.57\, V$,then the emf of the cell at $298\, K$ is $......\, V$. (in $.5$)

$EMF$ of the following concentration cell will be ............. $V$$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

Consider the cell whose $emf$ is $1.01 \ V$. $Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$ What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$

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$EMF$ of the following concentration cell will be ............. $V$
$Cl_2 \ (1 \ atm) | Cl^{-} \ (1 \ M) || Cl^{-} \ (0.1 \ M) | Cl_2 \ (2 \ atm)$

Consider the cell whose $emf$ is $1.01 \ V$.
$Pt, H_2(1 \ atm) | H^{+}(pH = 4) || Ag^{+}(xM) | Ag$
What is the value of $x$? (Given: $E^o_{Ag^{+}|Ag} = +0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06$)

In a cell,a copper electrode was used as a cathode. What is the electrode potential (in $V$) of the copper electrode dipped in $0.1 \ M \ Cu^{2+}$ solution at $298 \ K$?
$(E_{Cu^{2+}/Cu}^{\ominus} = 0.34 \ V; \frac{2.303 \ RT}{F} = 0.06 \ V)$