Consider the electrochemical cell shown in the figure where a metal electrode $(M)$ undergoes a redox reaction by forming $M^{+}$ $(M \rightarrow M^{+} + e^{-})$. The cation $M^{+}$ is present in two different concentrations $c_{1}$ and $c_{2}$. Which of the following statements is correct for generating a positive cell potential?
- AIf $c_{1}$ is present at the anode,then $c_{1} = c_{2}$
- BIf $c_{1}$ is present at the cathode,then $c_{1} < c_{2}$
- CIf $c_{1}$ is present at the cathode,then $c_{1} > c_{2}$
- DIf $c_{1}$ is present at the anode,then $c_{1} > c_{2}$
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The cell,$Zn\ |\ Zn^{2+} \,(1\ M)\ ||\ Cu^{2+}\ (1\ M)\ |\ Cu$ $(E^o_{cell} = 1.10\ V)$ was allowed to be completely discharged at $298\ K.$ The relative concentration of $Zn^{2+}$ to $Cu^{2+}$ $\left( \frac{[Zn^{2+}]}{[Cu^{2+}]} \right)$ is
At $298 \, K$,the standard reduction potential for $Cu^{2+}/Cu$ electrode is $0.34 \, V$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.
DifficultJEE MAIN 2023
View SolutionAt $298 \ K$,if the $emf$ of the cell corresponding to the reaction,$Zn_{(s)} + 2H^+_{(aq)} \rightarrow Zn^{2+}(0.01 \ M) + H_{2(g)}(1 \ atm)$ is $0.28 \ V$,then the $pH$ of the solution at the hydrogen electrode is (Given: $\frac{2.303 \ RT}{F} = 0.06 \ V$,$E^o_{Zn^{2+}|Zn} = -0.76 \ V$)
$Pt_{(s)} | H_{2(g)} (1 \ atm) | H^{+} (pH = 2) || H^{+} (pH = 3) | H_{2(g)} (1 \ atm) | Pt_{(s)}$ cell reaction will be
Easy
View SolutionAt $298 \ K$,some standard electrode potentials are given below:
$Pb^{2+} / Pb$ $-0.13 \ V$ $Ni^{2+} / Ni$ $-0.24 \ V$ $Cd^{2+} / Cd$ $-0.40 \ V$ $Fe^{2+} / Fe$ $-0.44 \ V$
Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$
| $Pb^{2+} / Pb$ | $-0.13 \ V$ |
| $Ni^{2+} / Ni$ | $-0.24 \ V$ |
| $Cd^{2+} / Cd$ | $-0.40 \ V$ |
| $Fe^{2+} / Fe$ | $-0.44 \ V$ |
Metal rods $X$ and $Y$ are inserted into a solution containing $0.001 \ M$ $X^{2+}$ and $0.1 \ M$ $Y^{2+}$ at $298 \ K$ and connected by a conducting wire. This results in the dissolution of $X$. The correct combination$(s)$ of $X$ and $Y$ are,respectively:
(Given: Gas constant,$R = 8.314 \ J \ K^{-1} \ mol^{-1}$,Faraday constant,$F = 96500 \ C \ mol^{-1}$)
$(A) \ Cd$ and $Ni \ \ (B) \ Cd$ and $Fe \ \ (C) \ Ni$ and $Pb \ \ (D) \ Ni$ and $Fe$
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