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Question 14 Marks
(a) Out of the following pairs, predict with reason which pair will allow greater conduction electricity:
(i) Silver wire at $30^{\circ}C$ or silver wire at $60^{\circ}C.$
(ii) 0.1 M $CH_{3}COOH$ solution or 1 M $CH_{3}COOH$ solution.
(iii) KCl solution at $20^{\circ}C$ or KCl solution at $50^{\circ}C$.
(b) Give two points of differences between electrochemical and electrolytic cells.
Answer
(a) (i) Silver wire at $30^{\circ} C$ because as temperature increases, resistance increases so conduction increases.
(ii) $0.1 M CH _3 COOH$, because on dilution degree of ionization increases hence conduction increases.
(iii) KCl solution at $50^{\circ} C$ because at high temperature mobility of ions increases hence conduction increases.
(b) Electrochemical CellElectrolytic Cell
(1) Anode : - Ve
Cathode : + Ve
Anode : + Ve
Cathode : - Ve
(2) It converts chemical energy into electrical energyIt converts electrical energy into chemical energy.
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Question 24 Marks
Write the cell reaction and calculate the emf of the following cell at 298 K
$S n _{( s )}\left| S n ^{2+}( 0 . 0 0 4 ~ M ) \| H ^{+}( 0 . 0 2 0 ~ M )\right| H _2( g ) \overline{( 1 )} \mid P t s$ (Given : $ES _{ n }{ }^{2+} / Sn =- 0 . 1 4 V$ )
Answer
Cell reaction:
At anode: $Sn \longrightarrow Sn^{2+} + 2e^{-} (OX)$
At cathode: $2H^{+} + 2e^{-} \longrightarrow H_{2} (Red)$
$Sn + 2H^{+} \longrightarrow Sn^{2+} + H_{2} (Redox)$
$E_{cell}^{0} = E_{SHE} - E_{RHE}$
$= 0.0 - (-0.14) = 0.14 $
$E = E_{cell}^{0} - \frac{0.0591}{2} log \frac{[Sn^{2+}](P_{H_{2}})}{[H^{+}]^{2}}$
$E = 0.14 - \frac{0.0591}{2} log \frac{0.004 \times 1}{(0.02)^{2}}$
$= 0.14 - 0.029 log \frac{4 \times 10^{-3}}{2 \times 2 \times 10^{-4}}$
$= 0.14 - 0.029 log 10$
$E = 0.111 V$
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Question 34 Marks
What is the difference between EMF and Cell potential (potential difference)?
Answer
S. No.EMFCell Potential (Potential difference)
1.EMF is the potential difference between two electrodes when no current is flowing in the circuit.Cell potential measures the difference in the potentials of the two half-cells when electric current flows through the cell.
2.EMF is the maximum voltage which can be obtained from the cell.Cell potential is always less than the maximum voltage obtainable from the cell.
3.EMF corresponds to the maximum useful work obtained from the galvanic cell.Cell potential does not correspond to the maximum useful work.
4.EMF is measured by potentiometerCell potential is measured by using a voltmeter.
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4 Marks Questions - Chemistry STD 12 Science Questions - Vidyadip