(B) The given reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$. In this reaction,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to

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(B) The given reaction is $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$. In this reaction,$Zn$ is oxidized to $Zn^{2+}$ (Anode) and $Fe^{2+}$ is reduced to $Fe$ (Cathode). The standard cell potential is calculated using the formula: $E^{0}_{cell} = E^{0}_{Oxi}(\text{anode}) + E^{0}_{Red}(\text{cathode})$. Since $E^{0}_{Oxi}(\text{Fe/Fe}^{2+}) = +0.41 \ V$,the reduction potential $E^{0}_{Red}(\text{Fe}^{2+}\text{/Fe}) = -0.41 \ V$. Thus,$E^{0}_{cell} = E^{0}_{Oxi}(\text{Zn/Zn}^{2+}) + E^{0}_{Red}(\text{Fe}^{2+}\text{/Fe}) = 0.76 \ V + (-0.41 \ V) = +0.35 \ V$.

Class 12 Chemistry Electrochemistry Electrode potential and ECell

Medium

For the half-reactions $Zn \rightarrow Zn^{2+} + 2e^{-}$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$,the standard oxidation potentials are given as $E^{0}_{Oxi} = +0.76 \ V$ and $E^{0}_{Oxi} = +0.41 \ V$ respectively. Calculate the cell potential for the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ in $V$.

A
$-0.35$
B
$+0.35$
C
$1.10$
D
$-1.10$

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Class 12

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Electrochemistry

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Electrode potential and ECell

At $25^\circ C$,$1 \, atm$ of gas $X$ is bubbled through a solution containing $1 \, M$ $Y^-$ and $1 \, M$ $Z^-$. If the reduction potential order is $Z > Y > X$,then:

Easy

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The oxidation potentials of the following half-cell reactions are given:
$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$
What will be the $EMF$ of the cell whose cell reaction is:
$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$

Medium

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Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

MediumMHT CET 2010

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The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

DifficultMHT CET 2026

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When a rod of metal $A$ is dipped in an aqueous solution of metal $B$ (concentration of $B^{2+}$ ion being $1 \ M$) at $25 \ ^oC$,the standard electrode potentials are $E^o_{A^{2+}/A} = -0.76 \ V$ and $E^o_{B^{2+}/B} = +0.34 \ V$. What will happen?

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At $25^\circ C$,$1 \, atm$ of gas $X$ is bubbled through a solution containing $1 \, M$ $Y^-$ and $1 \, M$ $Z^-$. If the reduction potential order is $Z > Y > X$,then:

The oxidation potentials of the following half-cell reactions are given:$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$What will be the $EMF$ of the cell whose cell reaction is:$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$

Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

The $E^0_{cell}$ of the cell $Al(s)|Al^{3+}(1M)||Pb^{2+}(1M)|Pb(s)$ is $1.5 \text{ V}$. If $E^0_{Pb^{2+}/Pb}$ is $-0.14 \text{ V}$, then the standard electrode potential $E^0_{Al^{3+}/Al}$ will be: (in $\text{ V}$)

When a rod of metal $A$ is dipped in an aqueous solution of metal $B$ (concentration of $B^{2+}$ ion being $1 \ M$) at $25 \ ^oC$,the standard electrode potentials are $E^o_{A^{2+}/A} = -0.76 \ V$ and $E^o_{B^{2+}/B} = +0.34 \ V$. What will happen?

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The oxidation potentials of the following half-cell reactions are given:
$Zn \to Zn^{2+} + 2e^-; E^o = 0.76 \, V$
$Fe \to Fe^{2+} + 2e^-; E^o = 0.44 \, V$
What will be the $EMF$ of the cell whose cell reaction is:
$Fe^{2+}_{(aq)} + Zn \to Zn^{2+}_{(aq)} + Fe$