285405
The amount of current in faraday required for the reduction of 1 mol of\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ions to \(\mathrm{Cr}^{3+}\) is
1 1 F
2 2 F
3 6 F
4 4 F
Explanation:
(c)\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)
6 F current will be required
Karnataka CET 2016
CHEMISTRY(KCET)
285406
How many coulombs of electricity are required for the oxidation of one mol of water to dioxygen?
1 \(1.93 \times 10^4 \mathrm{C}\)
2 \(19.3 \times 10^5 \mathrm{C}\)
3 \(9.65 \times 10^4 \mathrm{C}\)
4 \(1.93 \times 10^5 \mathrm{C}\)
Explanation:
(d)\(\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)
Hence coulombs of electricity required for the oxidation of one mole of water to dioxygen \(=2 \mathrm{~F}\)
\(=2 \times 96500=193000 \mathrm{C}=1.93 \times 10^5 \mathrm{C}\)
Karnataka CET 2015
CHEMISTRY(KCET)
285407
While charging the lead storage battery
1 \(\mathrm{PbSO}_4\) on cathode is reduced to Pb
2 \(\mathrm{PbSO}_4\) on anode is oxidized to \(\mathrm{PbO}_2\)
3 \(\mathrm{PbSO}_4\) on anode is reduced to Pb
4 \(\mathrm{PbSO}_4\) on cathode is oxidized to Pb .
285405
The amount of current in faraday required for the reduction of 1 mol of\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ions to \(\mathrm{Cr}^{3+}\) is
1 1 F
2 2 F
3 6 F
4 4 F
Explanation:
(c)\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)
6 F current will be required
Karnataka CET 2016
CHEMISTRY(KCET)
285406
How many coulombs of electricity are required for the oxidation of one mol of water to dioxygen?
1 \(1.93 \times 10^4 \mathrm{C}\)
2 \(19.3 \times 10^5 \mathrm{C}\)
3 \(9.65 \times 10^4 \mathrm{C}\)
4 \(1.93 \times 10^5 \mathrm{C}\)
Explanation:
(d)\(\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)
Hence coulombs of electricity required for the oxidation of one mole of water to dioxygen \(=2 \mathrm{~F}\)
\(=2 \times 96500=193000 \mathrm{C}=1.93 \times 10^5 \mathrm{C}\)
Karnataka CET 2015
CHEMISTRY(KCET)
285407
While charging the lead storage battery
1 \(\mathrm{PbSO}_4\) on cathode is reduced to Pb
2 \(\mathrm{PbSO}_4\) on anode is oxidized to \(\mathrm{PbO}_2\)
3 \(\mathrm{PbSO}_4\) on anode is reduced to Pb
4 \(\mathrm{PbSO}_4\) on cathode is oxidized to Pb .
285405
The amount of current in faraday required for the reduction of 1 mol of\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ions to \(\mathrm{Cr}^{3+}\) is
1 1 F
2 2 F
3 6 F
4 4 F
Explanation:
(c)\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)
6 F current will be required
Karnataka CET 2016
CHEMISTRY(KCET)
285406
How many coulombs of electricity are required for the oxidation of one mol of water to dioxygen?
1 \(1.93 \times 10^4 \mathrm{C}\)
2 \(19.3 \times 10^5 \mathrm{C}\)
3 \(9.65 \times 10^4 \mathrm{C}\)
4 \(1.93 \times 10^5 \mathrm{C}\)
Explanation:
(d)\(\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)
Hence coulombs of electricity required for the oxidation of one mole of water to dioxygen \(=2 \mathrm{~F}\)
\(=2 \times 96500=193000 \mathrm{C}=1.93 \times 10^5 \mathrm{C}\)
Karnataka CET 2015
CHEMISTRY(KCET)
285407
While charging the lead storage battery
1 \(\mathrm{PbSO}_4\) on cathode is reduced to Pb
2 \(\mathrm{PbSO}_4\) on anode is oxidized to \(\mathrm{PbO}_2\)
3 \(\mathrm{PbSO}_4\) on anode is reduced to Pb
4 \(\mathrm{PbSO}_4\) on cathode is oxidized to Pb .
285405
The amount of current in faraday required for the reduction of 1 mol of\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}\) ions to \(\mathrm{Cr}^{3+}\) is
1 1 F
2 2 F
3 6 F
4 4 F
Explanation:
(c)\(\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 e^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}\)
6 F current will be required
Karnataka CET 2016
CHEMISTRY(KCET)
285406
How many coulombs of electricity are required for the oxidation of one mol of water to dioxygen?
1 \(1.93 \times 10^4 \mathrm{C}\)
2 \(19.3 \times 10^5 \mathrm{C}\)
3 \(9.65 \times 10^4 \mathrm{C}\)
4 \(1.93 \times 10^5 \mathrm{C}\)
Explanation:
(d)\(\mathrm{H}_2 \mathrm{O} \rightarrow 2 \mathrm{H}^{+}+\frac{1}{2} \mathrm{O}_2+2 \mathrm{e}^{-}\)
Hence coulombs of electricity required for the oxidation of one mole of water to dioxygen \(=2 \mathrm{~F}\)
\(=2 \times 96500=193000 \mathrm{C}=1.93 \times 10^5 \mathrm{C}\)
Karnataka CET 2015
CHEMISTRY(KCET)
285407
While charging the lead storage battery
1 \(\mathrm{PbSO}_4\) on cathode is reduced to Pb
2 \(\mathrm{PbSO}_4\) on anode is oxidized to \(\mathrm{PbO}_2\)
3 \(\mathrm{PbSO}_4\) on anode is reduced to Pb
4 \(\mathrm{PbSO}_4\) on cathode is oxidized to Pb .
