The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. 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 relationship between ${E_1}$ and ${E_2}$?
- A${E_2} = 0 \neq {E_1}$
- B${E_1} > {E_2}$
- C${E_1} < {E_2}$
- D${E_1} = {E_2}$
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Similar Questions
$Emf$ of the following cell at $298 \,K$ in $V$ is $x \times 10^{-2}$. $Zn | Zn^{2+}(0.1 \,M) || Ag^{+}(0.01 \,M) | Ag$. The value of $x$ is .... . (Rounded off to the nearest integer) [Given: $E^{0}_{Zn^{2+}/Zn} = -0.76 \,V$; $E^{0}_{Ag^{+}/Ag} = +0.80 \,V$; $\frac{2.303 RT}{F} = 0.059$]
Calculate the $emf$ of the cell in which the following reaction takes place:
$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$
Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)
$Ni_{(s)} + 2Ag^{+}(0.002 \, M) \rightarrow Ni^{2+}(0.160 \, M) + 2Ag_{(s)}$
Given that $E^{\Theta}_{(cell)} = 1.05 \, V$ (in $, V$)
Medium
View SolutionFor a redox reaction,$Oxi + ne^- \rightarrow Red$,what is the correct form of the Nernst equation?
Medium
View SolutionAt $298 \, K$,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 SolutionThe electrode potential of the following half cell at $298 \ K$ is given by the cell reaction:
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.
$X | X^{2+}(0.001 \ M) || Y^{2+}(0.01 \ M) | Y$
The cell potential is $....... \times 10^{-2} \ V$ (Nearest integer).
Given: $E^0_{X^{2+} | X} = -2.36 \ V$,$E^0_{Y^{2+} | Y} = +0.36 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$.
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