(C) The cell reaction is: $M + Cu^{2+} \rightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0

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(C) The cell reaction is: $M + Cu^{2+} \rightarrow M^{2+} + Cu$ According to the Nernst equation: $E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.06}{n} \log \frac{[M^{2+}]}{[Cu^{2+}]}$ Given: $E^{\circ}_{\text{cell}} = 0.3 \ V$ $n = 2$ $[M^{2+}] = 0.1 \ M$ $E_{\text{cell}} = 0$ Substituting the values: $0 = 0.3 - \frac{0.06}{2} \log \frac{0.1}{[Cu^{2+}]}$ $0.3 = 0.03 \log \frac{0.1}{[Cu^{2+}]}$ $10 = \log \frac{0.1}{[Cu^{2+}]}$ $\frac{0.1}{[Cu^{2+}]} = 10^{10}$ $[Cu^{2+}] = \frac{0.1}{10^{10}} = 10^{-11} \ mol \ L^{-1}$

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

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A
$10^{-9}$
B
$10^{-8}$
C
$10^{-11}$
D
$10^{-10}$

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The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero? $\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$

Similar Questions

The relationship between standard reduction potential of a cell and the equilibrium constant is shown by:

Calculate the equilibrium constant for the cell reaction at $298 \ K$: $Cu_{(s)} + 2Ag^{+}_{(aq)} \rightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$,given $E^{0}_{cell} = 0.46 \ V$.

Considering the reaction $Cl_{2(g)} + 2Br^{-}_{(aq)} \rightarrow 2Cl^{-}_{(aq)} + Br_{2(l)}$,calculate the cell $emf$ in $V$ when $[Cl^{-}] = [Br_2] = [Br^{-}] = 0.01 \ M$ and $Cl_2$ gas is at $1 \ atm$ pressure. (Given $E^o = 0.29 \ V$ for the reaction)

For a reaction,$A_{(s)} + 2B_{(aq)}^{+} \rightleftharpoons A_{(aq)}^{2+} + 2B_{(s)}$,$K_{c}$ is $10^{12}$ at $25^{\circ} C$. The $E_{Cell}^{\circ}$ of the corresponding cell is $(F = 96500 \ C \ mol^{-1})$ (in $V$)

The change in electrode potential of $Cr^{3+}/Cr_2O_7^{2-}$ electrode at $25 \ ^oC$ due to change in $pH$ of its electrolytic solution from $1$ to $3$ is ........... $V$ (Assume: $[Cr_2O_7^{2-}]$ and $[Cr^{3+}]$ remain constant). Use: $\frac{2.303RT}{F} = 0.06$.

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The $E^{\circ}$ of $M \mid M^{2+} \parallel Cu^{2+} \mid Cu$ is $0.3 \ V$. At what concentration of $Cu^{2+}$ (in $mol \ L^{-1}$),the $E_{\text{cell}}$ value becomes zero?
$\left(\frac{2.303 \ RT}{F} = 0.06\right)$,$\left(\text{Conc. of } M^{2+} = 0.1 \ M\right)$