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Published on: 21/06/2021
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Questions + Answers key
Take MCQ Chemistry Test1.
The standard reduction potential of Fe3+, Fe2+ / Pt is +0.771 V. This half cell is connected with another half cell such that emf of the cell is 0.771V. What is the other half cell?
2.
State Ohms law.
3.
Corrosion of aluminum takes place at a much slower rate than iron. Give reason
4.
Explain the reactions taking place in the anode and cathode of a lead storage battery.
5.
Why does the emf of Leclanche cell decrease?
1.
(i) The emf of the standard hydrogen electrode is Zero volts.
Fe / Fe3+ Fe2+, H2(atm)pt E = 0.771 V
Emeasured = ER- EL
(ii) emf of SHE = 0
Emeasured = 0.771V - 0 = 0.771V.
(iii) ∴ The other half cell connected is Standard Hydrogen Electrode (SHE).
2.
This law can be stated as, "at constant temperature, the strength of the current flowing through a conductor is directly proportional to the potential difference and inversely proportional to the resistance of the conductor".
Thus, I = \(\frac{V}{R}\), V = RI, V = Volts, I = ampere R= ohms.
3.
Other metals such as aluminium, copper and silver also undergo corrosion, but at a slower rate than iron. Let us consider the reduction of aluminium,
\({ Al }_{ (s) }\rightarrow { Al }_{ (aq) }^{ 3+ }+{ 3e }^{ - }\)
Al3+, which reacts with oxygen in air to form a protective coating of Al2O3. This coating act as a protective film for the inner surface. So, further corrosion is prevented.
4.
Oxidation occurs at the anode
\({ Pb }_{ (s) }\rightarrow { Pb }_{ (aq) }^{ 2+ }+{ 2e }^{ - }\)
The Pb2+ions combine with \({ SO }_{ 4(aq) }^{2-}\) to from PbSO4 precipitate.
\({ Pb }_{ (aq) }^{ 2+ }+{ SO }_{ 4(aq) }^{ 2- }\rightarrow { PbSO }_{ 4(s) }\)
Reduction occurs at the cathode
\({ PbO }_{ 2(s) }+{ 4H }_{ (aq) }^{ + }+{ 2e }^{ - }\rightarrow { Pb }_{ (aq) }^{ 2+ }+{ 2H }_{ 2 }O(l)\)
The Pb2+ ions also combine with \({ SO }_{ 4(aq) }^{2-}\) ions from sulphuric acid to form PbSO4 precipitate.
\({ Pb }_{ (aq) }^{ 2+ }+{ SO }_{ 4(aq) }^{ -2 }\rightarrow { PbSO }_{ 4 }\)
5.
The overall redox reaction in Leclanche cell is
\( \mathrm{Zn}_{(\mathrm{s})}+2 \mathrm{NH}_{4 \text { (aq) }}^{+}+2 \mathrm{MnO}_{2(\mathrm{~s})} \longrightarrow \mathrm{Zn}_{(\mathrm{aq})}^{2+}+\mathrm{MnO}_{2} \mathrm{O}_{3(\mathrm{~s})}+\mathrm{H}_{2} \mathrm{O}(l)+2 \mathrm{NH}_{3} \)
The ammonia produced at the cathode combines with Zn2+ to form a complex ion [Zn (NH3)4]2+(aq). As the reaction proceeds the concentration of NH3 will decrease and the aqueous NH3 will increase which lead to the decrease in the emf of cell.
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