(C) The Nernst equation for the cell reaction is given by: $E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ Here,$n = 2$ (n

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(C) The Nernst equation for the cell reaction is given by: $E_{cell} = E_{cell}^o - \frac{0.0591}{n} \log \frac{[Zn^{2+}]}{[Cu^{2+}]}$ Here,$n = 2$ (number of electrons transferred). Substituting the given values: $E_{cell} = 1.10 - \frac{0.0591}{2} \log \frac{1}{0.1}$ $E_{cell} = 1.10 - 0.02955 \times \log(10)$ Since $\log(10) = 1$,we get: $E_{cell} = 1.10 - 0.02955 = 1.07045 \ V \approx 1.07 \ V$.

Class 12 Chemistry Electrochemistry Nernst equation and ECS

Medium

For the redox reaction $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$ taking place in a cell,$E_{cell}^o$ is $1.10 \ V$. $E_{cell}$ for the cell will be ............ $V$ $\left( 2.303 \frac{RT}{F} = 0.0591 \right)$

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A
$2.14$
B
$1.80$
C
$1.07$
D
$0.82$

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The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

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What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

MediumAP EAMCET 2025

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What is the potential of a half-cell consisting of a zinc electrode in $0.01 \ M$ $ZnSO_4$ solution at $25 \ ^\circ C$ (Given $E^o_{Zn^{2+}/Zn} = -0.763 \ V$) (in $V$)?

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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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For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

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For the redox reaction $Zn_{(s)} + Cu^{2+}(0.1 \ M) \to Zn^{2+}(1 \ M) + Cu_{(s)}$ taking place in a cell,$E_{cell}^o$ is $1.10 \ V$. $E_{cell}$ for the cell will be ............ $V$ $\left( 2.303 \frac{RT}{F} = 0.0591 \right)$

Similar Questions

The standard electrode potential for $Cu^{+2}/Cu$ is $0.34 \ V$. Calculate the reduction potential at $pH = 14$ for the above couple $V$ $[K_{sp}[Cu(OH)_2] = 1 \times 10^{-19}]$

What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

What is the potential of a half-cell consisting of a zinc electrode in $0.01 \ M$ $ZnSO_4$ solution at $25 \ ^\circ C$ (Given $E^o_{Zn^{2+}/Zn} = -0.763 \ V$) (in $V$)?

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

For the cell reaction $Sn_{(s)} + 2Ag^+_{(aq)} \rightarrow Sn^{2+}_{(aq)} + 2Ag_{(s)}$,what happens when the cell voltage increases?

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What is $E_{cell}$ (in $V$) of the following cell at $298 \ K$ ?
$(E^{\ominus}_{Zn^{2+}/Zn} = -0.76 \ V ; E^{\ominus}_{Ni^{2+}/Ni} = -0.25 \ V ; \frac{2.303 RT}{F} = 0.06 \ V)$
$Zn_{(s)} | Zn^{2+} (0.01 \ M) || Ni^{2+} (0.1 \ M) | Ni_{(s)}$

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