(A) The cell reaction is: $Fe^{2+} + Zn \to Zn^{2+} + Fe$.Here,$Zn$ is oxidized to $Zn^{2+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).$E^{\c

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(A) The cell reaction is: $Fe^{2+} + Zn \to Zn^{2+} + Fe$.Here,$Zn$ is oxidized to $Zn^{2+}$ (anode) and $Fe^{2+}$ is reduced to $Fe$ (cathode).$E^{\circ}_{cell} = E^{\circ}_{cathode} - E^{\circ}_{anode}$$E^{\circ}_{cell} = E^{\circ}_{Fe^{2+}/Fe} - E^{\circ}_{Zn^{2+}/Zn}$$E^{\circ}_{cell} = (-0.44 \, V) - (-0.76 \, V)$$E^{\circ}_{cell} = -0.44 + 0.76 = +0.32 \, V$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Medium

The standard electrode potentials of the half-cells are given below:
$Zn^{2+} + 2e^- \to Zn$; $E^{\circ} = -0.76 \, V$
$Fe^{2+} + 2e^- \to Fe$; $E^{\circ} = -0.44 \, V$
The $EMF$ of the cell $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

A
$+0.32$
B
$-0.32$
C
$+1.20$
D
$-1.20$

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Class 12

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Electrochemistry

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Electrode potential and ECell

What will happen when a silver ring is placed in an acidic solution?

Easy

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What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

Easy

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For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

Medium

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Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)

Medium

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Standard potential of electrode reaction $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ is $+0.34 \ V$. What is the standard potential of the reaction $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$?

MediumMHT CET 2025

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The standard electrode potentials of the half-cells are given below:$Zn^{2+} + 2e^- \to Zn$; $E^{\circ} = -0.76 \, V$$Fe^{2+} + 2e^- \to Fe$; $E^{\circ} = -0.44 \, V$The $EMF$ of the cell $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

Similar Questions

What will happen when a silver ring is placed in an acidic solution?

What will be the $EMF$ of the cell formed by the following half-cells in $V$?$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

For the cell reaction,$2Ce^{4+} + Co \to 2Ce^{3+} + Co^{2+}$,$E^\circ_{cell}$ is $1.89 \ V$. If $E^\circ_{Co^{2+}/Co} = -0.28 \ V$,then $E^\circ_{Ce^{4+}/Ce^{3+}}$ is equal to:

Calculate the cell potential in volts for the cell reaction given below:$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$Given:$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)

Standard potential of electrode reaction $Cu^{2+}_{(aq)} + 2e^{-} \rightarrow Cu_{(s)}$ is $+0.34 \ V$. What is the standard potential of the reaction $2Cu_{(s)} \rightarrow 2Cu^{2+}_{(aq)} + 4e^{-}$?

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The standard electrode potentials of the half-cells are given below:
$Zn^{2+} + 2e^- \to Zn$; $E^{\circ} = -0.76 \, V$
$Fe^{2+} + 2e^- \to Fe$; $E^{\circ} = -0.44 \, V$
The $EMF$ of the cell $Fe^{2+} + Zn \to Zn^{2+} + Fe$ is ............ $V$.

What will be the $EMF$ of the cell formed by the following half-cells in $V$?
$Mg^{2+} + 2e^- \to Mg_{(s)}; E = -2.37 \ V$
$Cu^{2+} + 2e^- \to Cu_{(s)}; E = +0.33 \ V$

Calculate the cell potential in volts for the cell reaction given below:
$Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$
Given:
$Zn \rightarrow Zn^{2+} + 2e^{-}, E^{0} = 0.76 \ V$
$Fe \rightarrow Fe^{2+} + 2e^{-}, E^{0} = 0.41 \ V$ (in $V$)