00. Electrode Potential
ELECTROCHEMISTRY

275772 In the electrochemical cell:
$\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(\mathbf{0 . 0 1} \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 . 0} \mathrm{M})\right \vert \mathrm{Cu}$, the emf of this Daniell cell is $E_{1}$. When the concentration of $\mathrm{ZnSO}_{4}$ is changed to $1.0 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ changed to $0.01 \mathrm{M}$, the emf changes to $E_{2}$. From the following which one is the relationship between $E_{1}$ and $E_{2}$ ? (Given, $\mathbf{R T} \mid \mathbf{F}=\mathbf{0 . 0 5 9}$ )

1 $\mathrm{E}_{1}<\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}$
JEE- 2017,JEE - 2003
ELECTROCHEMISTRY

275773 On the basis of the following $E^{0}$ values, the strongest oxidizing agent is
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.35 \mathrm{~V}$
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.77 \mathrm{~V}$

1 $\mathrm{Fe}^{3+}$
2 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$
3 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
4 $\mathrm{Fe}^{2+}$
ELECTROCHEMISTRY

275774 Zinc is used to protect iron from corrosion because

1 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {oxidation }}$ of iron
2 $\mathrm{E}_{\text {red }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {red }}$ of iron
3 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}=\mathrm{E}_{\text {oxidation }}$ of iron
4 Zinc is cheaper than iron
ELECTROCHEMISTRY

275778 The electrode potential of $\mathrm{M}^{2+} / \mathrm{M}$ for $3 \mathrm{~d}$ series elements shows positive value for :

1 $\mathrm{Fe}$
2 $\mathrm{Co}$
3 $\mathrm{Zn}$
4 $\mathrm{Cu}$
ELECTROCHEMISTRY

275772 In the electrochemical cell:
$\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(\mathbf{0 . 0 1} \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 . 0} \mathrm{M})\right \vert \mathrm{Cu}$, the emf of this Daniell cell is $E_{1}$. When the concentration of $\mathrm{ZnSO}_{4}$ is changed to $1.0 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ changed to $0.01 \mathrm{M}$, the emf changes to $E_{2}$. From the following which one is the relationship between $E_{1}$ and $E_{2}$ ? (Given, $\mathbf{R T} \mid \mathbf{F}=\mathbf{0 . 0 5 9}$ )

1 $\mathrm{E}_{1}<\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}$
JEE- 2017,JEE - 2003
ELECTROCHEMISTRY

275773 On the basis of the following $E^{0}$ values, the strongest oxidizing agent is
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.35 \mathrm{~V}$
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.77 \mathrm{~V}$

1 $\mathrm{Fe}^{3+}$
2 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$
3 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
4 $\mathrm{Fe}^{2+}$
ELECTROCHEMISTRY

275774 Zinc is used to protect iron from corrosion because

1 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {oxidation }}$ of iron
2 $\mathrm{E}_{\text {red }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {red }}$ of iron
3 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}=\mathrm{E}_{\text {oxidation }}$ of iron
4 Zinc is cheaper than iron
ELECTROCHEMISTRY

275778 The electrode potential of $\mathrm{M}^{2+} / \mathrm{M}$ for $3 \mathrm{~d}$ series elements shows positive value for :

1 $\mathrm{Fe}$
2 $\mathrm{Co}$
3 $\mathrm{Zn}$
4 $\mathrm{Cu}$
ELECTROCHEMISTRY

275772 In the electrochemical cell:
$\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(\mathbf{0 . 0 1} \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 . 0} \mathrm{M})\right \vert \mathrm{Cu}$, the emf of this Daniell cell is $E_{1}$. When the concentration of $\mathrm{ZnSO}_{4}$ is changed to $1.0 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ changed to $0.01 \mathrm{M}$, the emf changes to $E_{2}$. From the following which one is the relationship between $E_{1}$ and $E_{2}$ ? (Given, $\mathbf{R T} \mid \mathbf{F}=\mathbf{0 . 0 5 9}$ )

1 $\mathrm{E}_{1}<\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}$
JEE- 2017,JEE - 2003
ELECTROCHEMISTRY

275773 On the basis of the following $E^{0}$ values, the strongest oxidizing agent is
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.35 \mathrm{~V}$
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.77 \mathrm{~V}$

1 $\mathrm{Fe}^{3+}$
2 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$
3 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
4 $\mathrm{Fe}^{2+}$
ELECTROCHEMISTRY

275774 Zinc is used to protect iron from corrosion because

1 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {oxidation }}$ of iron
2 $\mathrm{E}_{\text {red }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {red }}$ of iron
3 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}=\mathrm{E}_{\text {oxidation }}$ of iron
4 Zinc is cheaper than iron
ELECTROCHEMISTRY

275778 The electrode potential of $\mathrm{M}^{2+} / \mathrm{M}$ for $3 \mathrm{~d}$ series elements shows positive value for :

1 $\mathrm{Fe}$
2 $\mathrm{Co}$
3 $\mathrm{Zn}$
4 $\mathrm{Cu}$
ELECTROCHEMISTRY

275772 In the electrochemical cell:
$\mathrm{Zn}\left \vert\mathrm{ZnSO}_{4}(\mathbf{0 . 0 1} \mathrm{M})\right \vert\left \vert\mathrm{CuSO}_{4}(\mathbf{1 . 0} \mathrm{M})\right \vert \mathrm{Cu}$, the emf of this Daniell cell is $E_{1}$. When the concentration of $\mathrm{ZnSO}_{4}$ is changed to $1.0 \mathrm{M}$ and that of $\mathrm{CuSO}_{4}$ changed to $0.01 \mathrm{M}$, the emf changes to $E_{2}$. From the following which one is the relationship between $E_{1}$ and $E_{2}$ ? (Given, $\mathbf{R T} \mid \mathbf{F}=\mathbf{0 . 0 5 9}$ )

1 $\mathrm{E}_{1}<\mathrm{E}_{2}$
2 $\mathrm{E}_{1}>\mathrm{E}_{2}$
3 $\mathrm{E}_{2}=0 \neq \mathrm{E}_{1}$
4 $\mathrm{E}_{1}=\mathrm{E}_{2}$
JEE- 2017,JEE - 2003
ELECTROCHEMISTRY

275773 On the basis of the following $E^{0}$ values, the strongest oxidizing agent is
$\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-} \rightarrow\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.35 \mathrm{~V}$
$\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ; \mathrm{E}^{0}=-0.77 \mathrm{~V}$

1 $\mathrm{Fe}^{3+}$
2 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{3-}$
3 $\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]^{4-}$
4 $\mathrm{Fe}^{2+}$
ELECTROCHEMISTRY

275774 Zinc is used to protect iron from corrosion because

1 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {oxidation }}$ of iron
2 $\mathrm{E}_{\text {red }}$ of $\mathrm{Zn}<\mathrm{E}_{\text {red }}$ of iron
3 $\mathrm{E}_{\text {oxidation }}$ of $\mathrm{Zn}=\mathrm{E}_{\text {oxidation }}$ of iron
4 Zinc is cheaper than iron
ELECTROCHEMISTRY

275778 The electrode potential of $\mathrm{M}^{2+} / \mathrm{M}$ for $3 \mathrm{~d}$ series elements shows positive value for :

1 $\mathrm{Fe}$
2 $\mathrm{Co}$
3 $\mathrm{Zn}$
4 $\mathrm{Cu}$