276241
Consider a fuel cell supplied with $1 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas and 10 moles of $\mathrm{O}_{2}$ gas. If fuel cell is operated at $96.5 \mathrm{~mA}$ current, how long will it deliver power? (Assume 1F $=96500 \mathrm{C} / \mathrm{mole}$ of electrons)
1 $1 \times 10^{6} \mathrm{~S}$
2 $0.5 \times 10^{6} \mathrm{~S}$
3 $2 \times 10^{6} \mathrm{~S}$
4 $4 \times 10^{6} \mathrm{~S}$
5 $5 \times 10^{6} \mathrm{~S}$
Explanation:
$\because \mathrm{w}=$ zit and $\mathrm{Z}=\frac{\text { At.wt }}{\mathrm{nF}}$ Since the cell is supplied with 1 mole of $\mathrm{H}_{2}$ gas and 10 mole of $\mathrm{O}_{2} \cdot \mathrm{H}_{2}$ is the limiting reagent, 1 mole of $\mathrm{H}_{2}$ reacts with 10 mole of $\mathrm{O}_{2}$. 1 mole of $\mathrm{H}_{2}$ required $0.5 \mathrm{~mol}$ of $\mathrm{O}_{2}$ charge requirement $=2 \mathrm{~F}$ $\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$ $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}, \mathrm{I}=96.5 \mathrm{~mA}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}=\frac{2 \times 96500}{9.65 \times 10^{-2}}=2 \times 10^{6} \mathrm{~S}$
Kerala-CEE-2017
ELECTROCHEMISTRY
276247
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
1 zinc is lighter than iron
2 zinc has lower melting point than iron
3 zinc has lower negative electrode potential than iron
4 zinc has higher negative electrode potential than iron
Explanation:
Reduction potential values of $\begin{aligned} & \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\mathrm{o}}=-0.44 \mathrm{~V} \end{aligned}$ Zinc can be coated on iron to produce galvanized iron but the reverse is not possible it is because Zinc has higher negative electrode potential than iron.
(NEET-II 2016)
ELECTROCHEMISTRY
276250
The standard electrode potential for Daniell cell is 1.1 volt. What is the standard Gibbs energy for the reaction?
1 $212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{E}^{\circ}=1.1$ volt. The standard electrode potential for Daniell cell is $\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}), \mathrm{n}=2 \\ & \therefore \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & \Delta \mathrm{G}^{\circ}=-2 \times 96500 \times 1.1 \\ & \Delta \mathrm{G}^{\circ}=-212.3 \mathrm{KJ} / \mathrm{mole} \end{aligned}$
276241
Consider a fuel cell supplied with $1 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas and 10 moles of $\mathrm{O}_{2}$ gas. If fuel cell is operated at $96.5 \mathrm{~mA}$ current, how long will it deliver power? (Assume 1F $=96500 \mathrm{C} / \mathrm{mole}$ of electrons)
1 $1 \times 10^{6} \mathrm{~S}$
2 $0.5 \times 10^{6} \mathrm{~S}$
3 $2 \times 10^{6} \mathrm{~S}$
4 $4 \times 10^{6} \mathrm{~S}$
5 $5 \times 10^{6} \mathrm{~S}$
Explanation:
$\because \mathrm{w}=$ zit and $\mathrm{Z}=\frac{\text { At.wt }}{\mathrm{nF}}$ Since the cell is supplied with 1 mole of $\mathrm{H}_{2}$ gas and 10 mole of $\mathrm{O}_{2} \cdot \mathrm{H}_{2}$ is the limiting reagent, 1 mole of $\mathrm{H}_{2}$ reacts with 10 mole of $\mathrm{O}_{2}$. 1 mole of $\mathrm{H}_{2}$ required $0.5 \mathrm{~mol}$ of $\mathrm{O}_{2}$ charge requirement $=2 \mathrm{~F}$ $\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$ $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}, \mathrm{I}=96.5 \mathrm{~mA}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}=\frac{2 \times 96500}{9.65 \times 10^{-2}}=2 \times 10^{6} \mathrm{~S}$
Kerala-CEE-2017
ELECTROCHEMISTRY
276247
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
1 zinc is lighter than iron
2 zinc has lower melting point than iron
3 zinc has lower negative electrode potential than iron
4 zinc has higher negative electrode potential than iron
Explanation:
Reduction potential values of $\begin{aligned} & \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\mathrm{o}}=-0.44 \mathrm{~V} \end{aligned}$ Zinc can be coated on iron to produce galvanized iron but the reverse is not possible it is because Zinc has higher negative electrode potential than iron.
(NEET-II 2016)
ELECTROCHEMISTRY
276250
The standard electrode potential for Daniell cell is 1.1 volt. What is the standard Gibbs energy for the reaction?
1 $212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{E}^{\circ}=1.1$ volt. The standard electrode potential for Daniell cell is $\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}), \mathrm{n}=2 \\ & \therefore \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & \Delta \mathrm{G}^{\circ}=-2 \times 96500 \times 1.1 \\ & \Delta \mathrm{G}^{\circ}=-212.3 \mathrm{KJ} / \mathrm{mole} \end{aligned}$
276241
Consider a fuel cell supplied with $1 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas and 10 moles of $\mathrm{O}_{2}$ gas. If fuel cell is operated at $96.5 \mathrm{~mA}$ current, how long will it deliver power? (Assume 1F $=96500 \mathrm{C} / \mathrm{mole}$ of electrons)
1 $1 \times 10^{6} \mathrm{~S}$
2 $0.5 \times 10^{6} \mathrm{~S}$
3 $2 \times 10^{6} \mathrm{~S}$
4 $4 \times 10^{6} \mathrm{~S}$
5 $5 \times 10^{6} \mathrm{~S}$
Explanation:
$\because \mathrm{w}=$ zit and $\mathrm{Z}=\frac{\text { At.wt }}{\mathrm{nF}}$ Since the cell is supplied with 1 mole of $\mathrm{H}_{2}$ gas and 10 mole of $\mathrm{O}_{2} \cdot \mathrm{H}_{2}$ is the limiting reagent, 1 mole of $\mathrm{H}_{2}$ reacts with 10 mole of $\mathrm{O}_{2}$. 1 mole of $\mathrm{H}_{2}$ required $0.5 \mathrm{~mol}$ of $\mathrm{O}_{2}$ charge requirement $=2 \mathrm{~F}$ $\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$ $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}, \mathrm{I}=96.5 \mathrm{~mA}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}=\frac{2 \times 96500}{9.65 \times 10^{-2}}=2 \times 10^{6} \mathrm{~S}$
Kerala-CEE-2017
ELECTROCHEMISTRY
276247
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
1 zinc is lighter than iron
2 zinc has lower melting point than iron
3 zinc has lower negative electrode potential than iron
4 zinc has higher negative electrode potential than iron
Explanation:
Reduction potential values of $\begin{aligned} & \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\mathrm{o}}=-0.44 \mathrm{~V} \end{aligned}$ Zinc can be coated on iron to produce galvanized iron but the reverse is not possible it is because Zinc has higher negative electrode potential than iron.
(NEET-II 2016)
ELECTROCHEMISTRY
276250
The standard electrode potential for Daniell cell is 1.1 volt. What is the standard Gibbs energy for the reaction?
1 $212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{E}^{\circ}=1.1$ volt. The standard electrode potential for Daniell cell is $\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}), \mathrm{n}=2 \\ & \therefore \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & \Delta \mathrm{G}^{\circ}=-2 \times 96500 \times 1.1 \\ & \Delta \mathrm{G}^{\circ}=-212.3 \mathrm{KJ} / \mathrm{mole} \end{aligned}$
276241
Consider a fuel cell supplied with $1 \mathrm{~mol}$ of $\mathrm{H}_{2}$ gas and 10 moles of $\mathrm{O}_{2}$ gas. If fuel cell is operated at $96.5 \mathrm{~mA}$ current, how long will it deliver power? (Assume 1F $=96500 \mathrm{C} / \mathrm{mole}$ of electrons)
1 $1 \times 10^{6} \mathrm{~S}$
2 $0.5 \times 10^{6} \mathrm{~S}$
3 $2 \times 10^{6} \mathrm{~S}$
4 $4 \times 10^{6} \mathrm{~S}$
5 $5 \times 10^{6} \mathrm{~S}$
Explanation:
$\because \mathrm{w}=$ zit and $\mathrm{Z}=\frac{\text { At.wt }}{\mathrm{nF}}$ Since the cell is supplied with 1 mole of $\mathrm{H}_{2}$ gas and 10 mole of $\mathrm{O}_{2} \cdot \mathrm{H}_{2}$ is the limiting reagent, 1 mole of $\mathrm{H}_{2}$ reacts with 10 mole of $\mathrm{O}_{2}$. 1 mole of $\mathrm{H}_{2}$ required $0.5 \mathrm{~mol}$ of $\mathrm{O}_{2}$ charge requirement $=2 \mathrm{~F}$ $\mathrm{H}_{2} \rightarrow 2 \mathrm{H}^{+}+2 \mathrm{e}^{-}$ $\mathrm{I}=\frac{\mathrm{Q}}{\mathrm{t}}, \mathrm{I}=96.5 \mathrm{~mA}$ $\mathrm{t}=\frac{\mathrm{Q}}{\mathrm{I}}=\frac{2 \times 96500}{9.65 \times 10^{-2}}=2 \times 10^{6} \mathrm{~S}$
Kerala-CEE-2017
ELECTROCHEMISTRY
276247
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
1 zinc is lighter than iron
2 zinc has lower melting point than iron
3 zinc has lower negative electrode potential than iron
4 zinc has higher negative electrode potential than iron
Explanation:
Reduction potential values of $\begin{aligned} & \mathrm{E}_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\mathrm{o}}=-0.76 \mathrm{~V} \\ & \mathrm{E}_{\mathrm{Fe}^{2+} / \mathrm{Fe}}^{\mathrm{o}}=-0.44 \mathrm{~V} \end{aligned}$ Zinc can be coated on iron to produce galvanized iron but the reverse is not possible it is because Zinc has higher negative electrode potential than iron.
(NEET-II 2016)
ELECTROCHEMISTRY
276250
The standard electrode potential for Daniell cell is 1.1 volt. What is the standard Gibbs energy for the reaction?
1 $212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
2 $-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
3 $106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
4 $-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
Explanation:
Given that, $\mathrm{E}^{\circ}=1.1$ volt. The standard electrode potential for Daniell cell is $\begin{aligned} & \mathrm{Zn}(\mathrm{s})+\mathrm{Cu}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Zn}^{2+}(\mathrm{aq})+\mathrm{Cu}(\mathrm{s}), \mathrm{n}=2 \\ & \therefore \Delta \mathrm{G}^{\circ}=-\mathrm{nFE}^{\circ} \\ & \Delta \mathrm{G}^{\circ}=-2 \times 96500 \times 1.1 \\ & \Delta \mathrm{G}^{\circ}=-212.3 \mathrm{KJ} / \mathrm{mole} \end{aligned}$
