(C) The Nernst equation for the electrode reaction $Cu^{2+} + 2e^- \rightarrow Cu(s)$ is:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \f

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(C) The Nernst equation for the electrode reaction $Cu^{2+} + 2e^- \rightarrow Cu(s)$ is:$E_{Cu^{2+}/Cu} = E^0_{Cu^{2+}/Cu} - \frac{0.0591}{n} \log \frac{1}{[Cu^{2+}]}$Here, $n = 2$ and $[Cu^{2+}] = 0.1 \text{ M} = 10^{-1} \text{ M}$.Substituting the values:$E = 0.34 - \frac{0.0591}{2} \log \frac{1}{10^{-1}}$$E = 0.34 - 0.02955 \log(10)$Since $\log(10) = 1$, we get:$E = 0.34 - 0.02955 = 0.31045 \text{ V} \approx 0.31 \text{ V}$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Difficult

The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

MHT CET 2026 All MHT CET

A
$0.64$
B
$0.34$
C
$0.31$
D
$0.37$

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Class 12

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Electrochemistry

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Nernst equation and ECS

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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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For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.
Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

Medium

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The e.m.f. of the cell $Zn | Zn^{2+} (0.01 \ M) || Fe^{2+} (0.001 \ M) | Fe$ at $298 \ K$ is $0.2905 \ V$. The value of the equilibrium constant for the cell reaction is:

DifficultIIT 2004

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Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.

EasyKCET 2023

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Give the $E_{cell}$ equation for the general reaction: $aA + bB \to cC + dD$.

Easy

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$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$	$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$	$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

The standard electrode potential of $Cu^{2+}/Cu$ is $0.34 \text{ V}$ at $298 \text{ K}$. Calculate its electrode potential at the same temperature when the $Cu^{2+}$ ion concentration is $0.1 \text{ M}$. (in $\text{ V}$)

Similar Questions

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

For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

The e.m.f. of the cell $Zn | Zn^{2+} (0.01 \ M) || Fe^{2+} (0.001 \ M) | Fe$ at $298 \ K$ is $0.2905 \ V$. The value of the equilibrium constant for the cell reaction is:

Consider the following $4$ electrodes:$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$$B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$$D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$Arrange the reduction potential of these electrodes in decreasing order.

Give the $E_{cell}$ equation for the general reaction: $aA + bB \to cC + dD$.

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For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.
Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

Consider the following $4$ electrodes:
$A$. $Ag^{+}(0.0001 \ M) / Ag_{(s)}$ $B$. $Ag^{+}(0.1 \ M) / Ag_{(s)}$
$C$. $Ag^{+}(0.01 \ M) / Ag_{(s)}$ $D$. $Ag^{+}(0.001 \ M) / Ag_{(s)}$

$E^{\circ}_{Ag^{+} / Ag} = +0.80 \ V$
Arrange the reduction potential of these electrodes in decreasing order.