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(D) The Nernst equation for the given cell reaction is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.Here,the num

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less than $E^{\circ}_{cell}$ by $0.0296 \ V$

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(D) The Nernst equation for the given cell reaction is: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{n} \log \frac{[Cd^{2+}]}{[Cu^{2+}]}$.Here,the number of electrons transferred $n = 2$.Given that $[Cd^{2+}] = 10 [Cu^{2+}]$,so the ratio $\frac{[Cd^{2+}]}{[Cu^{2+}]} = 10$.Substituting these values into the equation: $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} \log(10)$.Since $\log(10) = 1$,we get $E_{cell} = E^{\circ}_{cell} - \frac{0.0591}{2} 	imes 1$.$E_{cell} = E^{\circ}_{cell} - 0.02955 \ V \approx E^{\circ}_{cell} - 0.0296 \ V$.Therefore,the $emf$ of the cell is less than $E^{\circ}_{cell}$ by $0.0296 \ V$.

Which of the following statements is correct regarding the $emf$ of the cell for the cell reaction,$Cd_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Cd^{2+}_{(aq)} + Cu_{(s)}$,if the concentration of $Cd^{2+}$ is $10$ times greater than the concentration of $Cu^{2+}_{(aq)}$ at $298 \ K$?

Similar Questions

If the standard electrode potential of $Cu^{2+}/Cu$ electrode is $0.34 \, V$,what is the electrode potential of $0.01 \, M$ concentration of $Cu^{2+}$ $(T = 298 \, K)$ (in $, V$)?

What will be the $EMF$ for the given cell?
$Pt | H_2(g, P_1) | H^{+}(aq.) (1 \ M) || H^{+}(aq.) (1 \ M) | H_2(g, P_2) | Pt$

Calculate the equilibrium constant of the reaction,$Cu_{(s)} + 2 Ag^{+}_{(aq)} \longrightarrow Cu^{2+}_{(aq)} + 2 Ag_{(s)}$,given that for the reaction $E^{\circ}_{cell} = 0.46 \ V$.

The $EMF$ of the cell $M | M^{n+} (0.02 \, M) || H^{+} (1 \, M) | H_{2(g)} (1 \, atm), Pt$ at $25 \, ^\circ C$ is $0.81 \, V$. Calculate the valency of the metal $(n)$ if the standard oxidation potential of the metal is $0.76 \, V$. (Use $\frac{2.303 \, RT}{F} = 0.06, \log \, 2 = 0.3$)

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