285397
Given :\(E_{\mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+2}}^{\circ}=1.5 \mathrm{~V}\) and \(E_{\mathrm{Mn}^{+4} \mid \mathrm{Mn}^{+2}}^{\circ}=1.2 \mathrm{~V}\), then \(E^{\circ} \mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+4}\) is
1 0.3 V
2 0.1 V
3 1.7 V
4 2.1 V
Explanation:
(c)\(\mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.5 \mathrm{~V}\)
\(\begin{aligned}
& \Delta G_1^{\circ}=-n F E^{\circ}=-5 \times F \times 1.5 \Rightarrow-7.5 \mathrm{~F} \\
& \mathrm{Mn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.2 \mathrm{~V} \\
& \Delta G_2^{\circ}=-n F E^{\circ}=-2 \times F \times 1.2 \Rightarrow-2.4 \mathrm{~F}
\end{aligned}\)
On subtracting eqn (ii) from (i), we get,
\(\mathrm{Mn}^{7+}+3 e^{-} \rightarrow \mathrm{Mn}^{4+} ; E^{\circ}=\) ?
\(\Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)
\(-3 F E^{\circ}=-7.5 \mathrm{~F}-(-2.4 \mathrm{~F})\)
\(\Rightarrow-3 F E^{\circ}=-5.1 \mathrm{~F} \therefore E^{\circ}=1.7 \mathrm{~V}\)
Karnataka CET 2019
CHEMISTRY(KCET)
285398
The charge required for the reduction of 1 mole of\(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\) is
1 1 F
2 3 F
3 5 F
4 7 F
Explanation:
(b)\(\stackrel{+7}{\mathrm{MnO}_4^{-}}+3 \mathrm{e}^{-} \xrightarrow{+4} \mathrm{MnO}_2\)
3 F charge will be required to reduce 1 mole of \(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\).
Karnataka CET 2018
CHEMISTRY(KCET)
285399
At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is
285397
Given :\(E_{\mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+2}}^{\circ}=1.5 \mathrm{~V}\) and \(E_{\mathrm{Mn}^{+4} \mid \mathrm{Mn}^{+2}}^{\circ}=1.2 \mathrm{~V}\), then \(E^{\circ} \mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+4}\) is
1 0.3 V
2 0.1 V
3 1.7 V
4 2.1 V
Explanation:
(c)\(\mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.5 \mathrm{~V}\)
\(\begin{aligned}
& \Delta G_1^{\circ}=-n F E^{\circ}=-5 \times F \times 1.5 \Rightarrow-7.5 \mathrm{~F} \\
& \mathrm{Mn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.2 \mathrm{~V} \\
& \Delta G_2^{\circ}=-n F E^{\circ}=-2 \times F \times 1.2 \Rightarrow-2.4 \mathrm{~F}
\end{aligned}\)
On subtracting eqn (ii) from (i), we get,
\(\mathrm{Mn}^{7+}+3 e^{-} \rightarrow \mathrm{Mn}^{4+} ; E^{\circ}=\) ?
\(\Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)
\(-3 F E^{\circ}=-7.5 \mathrm{~F}-(-2.4 \mathrm{~F})\)
\(\Rightarrow-3 F E^{\circ}=-5.1 \mathrm{~F} \therefore E^{\circ}=1.7 \mathrm{~V}\)
Karnataka CET 2019
CHEMISTRY(KCET)
285398
The charge required for the reduction of 1 mole of\(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\) is
1 1 F
2 3 F
3 5 F
4 7 F
Explanation:
(b)\(\stackrel{+7}{\mathrm{MnO}_4^{-}}+3 \mathrm{e}^{-} \xrightarrow{+4} \mathrm{MnO}_2\)
3 F charge will be required to reduce 1 mole of \(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\).
Karnataka CET 2018
CHEMISTRY(KCET)
285399
At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is
285397
Given :\(E_{\mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+2}}^{\circ}=1.5 \mathrm{~V}\) and \(E_{\mathrm{Mn}^{+4} \mid \mathrm{Mn}^{+2}}^{\circ}=1.2 \mathrm{~V}\), then \(E^{\circ} \mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+4}\) is
1 0.3 V
2 0.1 V
3 1.7 V
4 2.1 V
Explanation:
(c)\(\mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.5 \mathrm{~V}\)
\(\begin{aligned}
& \Delta G_1^{\circ}=-n F E^{\circ}=-5 \times F \times 1.5 \Rightarrow-7.5 \mathrm{~F} \\
& \mathrm{Mn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.2 \mathrm{~V} \\
& \Delta G_2^{\circ}=-n F E^{\circ}=-2 \times F \times 1.2 \Rightarrow-2.4 \mathrm{~F}
\end{aligned}\)
On subtracting eqn (ii) from (i), we get,
\(\mathrm{Mn}^{7+}+3 e^{-} \rightarrow \mathrm{Mn}^{4+} ; E^{\circ}=\) ?
\(\Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)
\(-3 F E^{\circ}=-7.5 \mathrm{~F}-(-2.4 \mathrm{~F})\)
\(\Rightarrow-3 F E^{\circ}=-5.1 \mathrm{~F} \therefore E^{\circ}=1.7 \mathrm{~V}\)
Karnataka CET 2019
CHEMISTRY(KCET)
285398
The charge required for the reduction of 1 mole of\(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\) is
1 1 F
2 3 F
3 5 F
4 7 F
Explanation:
(b)\(\stackrel{+7}{\mathrm{MnO}_4^{-}}+3 \mathrm{e}^{-} \xrightarrow{+4} \mathrm{MnO}_2\)
3 F charge will be required to reduce 1 mole of \(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\).
Karnataka CET 2018
CHEMISTRY(KCET)
285399
At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is
285397
Given :\(E_{\mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+2}}^{\circ}=1.5 \mathrm{~V}\) and \(E_{\mathrm{Mn}^{+4} \mid \mathrm{Mn}^{+2}}^{\circ}=1.2 \mathrm{~V}\), then \(E^{\circ} \mathrm{Mn}^{+7} \mid \mathrm{Mn}^{+4}\) is
1 0.3 V
2 0.1 V
3 1.7 V
4 2.1 V
Explanation:
(c)\(\mathrm{Mn}^{7+}+5 e^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.5 \mathrm{~V}\)
\(\begin{aligned}
& \Delta G_1^{\circ}=-n F E^{\circ}=-5 \times F \times 1.5 \Rightarrow-7.5 \mathrm{~F} \\
& \mathrm{Mn}^{4+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Mn}^{2+} ; E^{\circ}=1.2 \mathrm{~V} \\
& \Delta G_2^{\circ}=-n F E^{\circ}=-2 \times F \times 1.2 \Rightarrow-2.4 \mathrm{~F}
\end{aligned}\)
On subtracting eqn (ii) from (i), we get,
\(\mathrm{Mn}^{7+}+3 e^{-} \rightarrow \mathrm{Mn}^{4+} ; E^{\circ}=\) ?
\(\Delta G_3^{\circ}=\Delta G_1^{\circ}-\Delta G_2^{\circ}\)
\(-3 F E^{\circ}=-7.5 \mathrm{~F}-(-2.4 \mathrm{~F})\)
\(\Rightarrow-3 F E^{\circ}=-5.1 \mathrm{~F} \therefore E^{\circ}=1.7 \mathrm{~V}\)
Karnataka CET 2019
CHEMISTRY(KCET)
285398
The charge required for the reduction of 1 mole of\(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\) is
1 1 F
2 3 F
3 5 F
4 7 F
Explanation:
(b)\(\stackrel{+7}{\mathrm{MnO}_4^{-}}+3 \mathrm{e}^{-} \xrightarrow{+4} \mathrm{MnO}_2\)
3 F charge will be required to reduce 1 mole of \(\mathrm{MnO}_4^{-}\)to \(\mathrm{MnO}_2\).
Karnataka CET 2018
CHEMISTRY(KCET)
285399
At a particular temperature, the ratio of molar conductance to specific conductance of 0.01 M NaCl solution is
