(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E = E^0 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[

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(A) The cell reaction is: $Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)$.Using the Nernst equation: $E = E^0 - \frac{0.0592}{2} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$.For $E_1$: $[Zn^{2+}] = 0.01 \ M$ and $[Cu^{2+}] = 1 \ M$,so $E_1 = E^0 - \frac{0.0592}{2} \log(0.01) = E^0 - \frac{0.0592}{2} (-2) = E^0 + 0.0592$.For $E_2$: $[Zn^{2+}] = 1 \ M$ and $[Cu^{2+}] = 0.01 \ M$,so $E_2 = E^0 - \frac{0.0592}{2} \log(100) = E^0 - \frac{0.0592}{2} (2) = E^0 - 0.0592$.Comparing the two,$E_1 > E_2$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

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For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

A
$E_1 > E_2$
B
$E_1 < E_2$
C
$E_1 = E_2$
D
$E_2 = 0 \neq E_1$

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Nernst equation and ECS

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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If a hydrogen electrode is dipped in two solutions of $pH = 3$ and $pH = 6$ and a salt bridge is connected,the e.m.f. of the resulting cell is ............ $V$.

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The cell reaction of the given cell is spontaneous if:
$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$

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The standard $emf$ of a galvanic cell involving $3$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the reaction of the cell is:

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When is the reduction potential of a hydrogen half-cell negative?

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For a $Daniel$ cell $Zn | ZnSO_{4(0.01 \ M)} || CuSO_{4(1 \ M)} | Cu$ at $298 \ K$,the cell potential is $E_1$. When the concentrations of $ZnSO_4$ and $CuSO_4$ are changed to $1 \ M$ and $0.01 \ M$ respectively,the cell potential becomes $E_2$. Determine the relationship between $E_1$ and $E_2$.

Similar Questions

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: $\frac{2.303RT}{F} = 0.06$.

If a hydrogen electrode is dipped in two solutions of $pH = 3$ and $pH = 6$ and a salt bridge is connected,the e.m.f. of the resulting cell is ............ $V$.

The cell reaction of the given cell is spontaneous if:$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$

The standard $emf$ of a galvanic cell involving $3$ moles of electrons in its redox reaction is $0.59 \ V$. The equilibrium constant for the reaction of the cell is:

When is the reduction potential of a hydrogen half-cell negative?

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The cell reaction of the given cell is spontaneous if:
$Pt | Cl_2 (P_1 \, atm) | Cl^{-} (1 \, M) || Cl^{-} (1 \, M) | Cl_2 (P_2 \, atm) | Pt$