(C) The cell reaction for the $Daniell$ cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Using the $Nernst$ equation:$E_{cell} = E_{cell

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(C) The cell reaction for the $Daniell$ cell is: $Zn(s) + Cu^{2+}(aq) \rightarrow Zn^{2+}(aq) + Cu(s)$.Using the $Nernst$ equation:$E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$For $E_1$:$E_1 = E^o - \frac{0.0591}{2} \log \frac{0.01}{1.0} = E^o - \frac{0.0591}{2} \log(10^{-2}) = E^o + 0.0591 \, V$.For $E_2$:$E_2 = E^o - \frac{0.0591}{2} \log \frac{1.0}{0.01} = E^o - \frac{0.0591}{2} \log(10^2) = E^o - 0.0591 \, V$.Comparing the two values,$E^o + 0.0591 > E^o - 0.0591$,therefore $E_1 > E_2$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:
$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

AIIMS 2008 All AIIMS

A
$E_1 = E_2$
B
$E_2 = 0 \neq E_1$
C
$E_1 > E_2$
D
$E_1 < E_2$

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Electrochemistry

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Nernst equation and ECS

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AIIMS 2008 Paper All AIIMS years →

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

MediumMHT CET 2025

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$A$ copper electrode is dipped in a $0.1 \, M$ copper sulfate solution at $25 \, ^\circ C$. Calculate the reduction potential of the copper electrode. $(E^o_{Cu^{2+}/Cu} = 0.34 \, V)$ (in $, V$)

Medium

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Calculate the $e.m.f.$ of the half-cell given below:
$Fe | FeSO_4$ $(a = 0.1 \ M)$
Given: $E^o_{OP} = 0.44 \ V$ (in $V$)

Medium

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What is the reduction electrode potential (in volts) of a copper electrode when $[Cu^{2+}]=0.01 \ M$ is in a solution at $25^{\circ} C$? $(E^{\circ}$ of $Cu^{2+}/Cu$ electrode is $+0.34 \ V)$

MediumTS EAMCET 2002

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What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction?
Given: $(\frac{2.303RT}{F} = 0.06)$
$Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$
Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$

Medium

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The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

Similar Questions

If $E^{\circ}(Mg^{+2}_{(aq)} \mid Mg_{(s)}) = -2.37 \ V$. What is the potential for $Mg_{(s)} \rightarrow Mg^{+2}_{(0.01 \ M)} + 2 \overline{e}$ at $298 \ K$?

$A$ copper electrode is dipped in a $0.1 \, M$ copper sulfate solution at $25 \, ^\circ C$. Calculate the reduction potential of the copper electrode. $(E^o_{Cu^{2+}/Cu} = 0.34 \, V)$ (in $, V$)

Calculate the $e.m.f.$ of the half-cell given below: $Fe | FeSO_4$ $(a = 0.1 \ M)$ Given: $E^o_{OP} = 0.44 \ V$ (in $V$)

What is the reduction electrode potential (in volts) of a copper electrode when $[Cu^{2+}]=0.01 \ M$ is in a solution at $25^{\circ} C$? $(E^{\circ}$ of $Cu^{2+}/Cu$ electrode is $+0.34 \ V)$

What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction? Given: $(\frac{2.303RT}{F} = 0.06)$ $Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$ Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$

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The $emf$ of a $Daniell$ cell at $298 \, K$ is $E_1$:
$Zn | ZnSO_4 \,(0.01 \, M) || CuSO_4 \,(1.0 \, M) | Cu$
When the concentration of $ZnSO_4$ is $1.0 \, M$ and that of $CuSO_4$ is $0.01 \, M$,the $emf$ changes to $E_2$. What is the relation between $E_1$ and $E_2$?

Calculate the $e.m.f.$ of the half-cell given below:
$Fe | FeSO_4$ $(a = 0.1 \ M)$
Given: $E^o_{OP} = 0.44 \ V$ (in $V$)

What minimum decomposition potential is necessary to produce $Cl_2$ gas in the following reaction?
Given: $(\frac{2.303RT}{F} = 0.06)$
$Sn^{+2} (1 \ M) + 2Cl^{-} (2 \ M) \rightleftharpoons Sn_{(s)} + Cl_2 (1 \ atm)$
Given: $E^{o}_{Sn^{+2}/Sn} = -0.14 \ V$,$E^{o}_{Cl_2/Cl^{-}} = 1.4 \ V$