(B) For the given cell reaction,the number of electrons transferred is $n = 2$. Using the Nernst equation at equilibrium: $E_{cell}^o = \frac{0.0591}{

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(B) For the given cell reaction,the number of electrons transferred is $n = 2$. Using the Nernst equation at equilibrium: $E_{cell}^o = \frac{0.0591}{n} \log K_c$. Substituting the values: $1.10 = \frac{0.0591}{2} \log K_c$. $\log K_c = \frac{1.10 \times 2}{0.0591} \approx 37.22$. Therefore,$K_c = 10^{37.22} \approx 10^{37}$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

The $E^o$ for the cell $Zn|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu$ is $1.10 \ V$ at $25 \ ^oC$. The equilibrium constant for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$ is of the order of:

AIPMT 1997 All AIPMT

A
$10^{-28}$
B
$10^{+37}$
C
$10^{+18}$
D
$10^{+17}$

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Class 12

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Electrochemistry

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Nernst equation and ECS

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AIPMT 1997 Paper All AIPMT years →

Which of the following conditions will increase the voltage of the cell,represented by the equation $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$?

EasyAIPMT 2001

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Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)

Difficult

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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

Difficult

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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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The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?

Medium

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The $E^o$ for the cell $Zn|Zn^{2+}_{(aq)}||Cu^{2+}_{(aq)}|Cu$ is $1.10 \ V$ at $25 \ ^oC$. The equilibrium constant for the reaction $Zn + Cu^{2+}_{(aq)} \rightleftharpoons Cu + Zn^{2+}_{(aq)}$ is of the order of:

Similar Questions

Which of the following conditions will increase the voltage of the cell,represented by the equation $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$?

Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$ (Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)

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

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 $.5$)

The cell reaction involving the quinhydrone electrode is given by the following equation:$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$What will be the electrode potential at $pH = 3$ (in $V$)?

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Calculate the $EMF$ of the cell: $Cr | Cr^{+3}(0.1 \, M) || Fe^{+2}(0.01 \, M) | Fe$
(Given: $E^o_{Cr^{+3}|Cr} = -0.75 \, V$,$E^o_{Fe^{+2}|Fe} = -0.45 \, V$) (in $, V$)

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

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?