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Q. No.The cell that will measure the standard electrode potential of a copper electrode is:
| 1. | Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, 1M) || Cu2+(aq, 1M) | Cu |
| 2. | Pt(s) | H2(g, 1 bar) | H+(aq, 1M) || Cu2+ (aq, 2M) | Cu |
| 3. | Pt(s) | H2(g, 1 bar) | H+(aq, 1M)|| Cu2+ (aq, 1M) | Cu |
| 4. | Pt(s) | H2(g, \(1 \over 10\) bar) | H+(aq, \(1 \over 10\)M) || Cu2+(aq, 1M) | Cu |
For the given cell, Mg | Mg2+ || Cu2+ | Cu
a. Mg acts as cathode
b. Cu acts as cathode
c. The cell reaction is \(Mg + Cu^{2+} \rightarrow Mg^{2+} + Cu\)
d. Cu is the oxidising agent
The correct choice among the given is -
| 1. | (a, b) | 2. | (b, c) |
| 3. | (c, d) | 4. | (a, d) |
The positive value of the standard electrode potential of Cu2+ / Cu indicates that-
| (a) | This redox couple is a stronger reducing agent than the H+ /H2 couple |
| (b) | This redox couple is a stronger oxidising agent than H+ /H2 |
| (c) | Cu can displace H2 from acid. |
| (d) | Cu cannot displace H2 from acid. |
1. (a, b)
2. (b, c)
3. (c, d)
4. (b, d)
What will happen during the electrolysis of aqueous solution of CuSO4 in the presence of Cu electrodes?
| (a) | Copper will deposit at cathode |
| (b) | Copper will dissolve at anode |
| (c) | Oxygen will be released at anode |
| (d) | Copper will deposit at anode |
The correct choice among the given is -
| 1. | (a, b) | 2. | (b, c) |
| 3. | (c, d) | 4. | (a, d) |
On electrolysis of dilute sulphuric acid using Platinum (Pt) electrode, the product obtained at the anode will be:
1. Oxygen gas
2. gas
3. gas
4. Hydrogen gas
Given below are two statements:
| Assertion (A): | Electrolysis of NaCl solution gives chlorine at anode instead of O2 . |
| Reason (R): | Formation of oxygen at anode requires overvoltage. |
| 1. | Both (A) and (R) are true and (R) is the correct explanation of (A). |
| 2. | Both (A) and (R) are true but (R) is not the correct explanation of (A). |
| 3. | (A) is true but (R) is false. |
| 4. | (A) is false but (R) is true. |
of some half cell reactions are given below.
| I: | \(\small\mathrm{H}^{+} (a q)+e^{-} \rightarrow \frac{1}{2} \mathrm{H}_2 (g )\quad ; \quad E_{\text {cell }}^0=0.00 \mathrm{~V} \) |
| II: | \(\small2 \mathrm{H}_2 \mathrm{O(l)} \rightarrow O_2 (g)+4 \mathrm{H}^{+} (a q)+4 e^{-} ; E_{\text {cell }}^{0}=1.23 \mathrm{~V} \) |
| III: | \(\small2 \mathrm{SO}_4^{2-} (a q) \rightarrow \mathrm{S}_2 \mathrm{O}_8^{2-} (a q)+2 e^{-}, E_{\text {cell }}^{0}=1.96 \mathrm{~V} \) |
The correct statements among the following are:
| a. | In dilute sulphuric acid solution, hydrogen will be reduced at cathode. |
| b. | In concentrated sulphuric acid solution, water will be oxidised at anode. |
| c. | In a dilute sulphuric acid solution, water will be oxidised at anode. |
| d. | In dilute sulphuric acid solution, SO42- ion will be oxidised to tetrathionate ion at anode. |
1. (a, b)
2. (b, c)
3. (c, d)
4. (a, c)
The half-cell reaction at the anode during the electrolysis of aqueous sodium chloride solution is represented by :
1. Na+(aq) + e- ⟶ Na(s) ; \(E_{cell}^{o} \ = \ -2.71 \ V \)
2. 2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e- ; \(E_{cell}^{o} \) = 1.23 V
3. H+(aq) + e- ⟶ \(\frac{1}{2}\)H2(g) ; \(E_{cell}^{o} \) = 0.00 V
4. Cl-(aq) ⟶ \(\frac{1}{2}\)Cl2(g) + e- ; \(E_{cell}^{o}\)
The correct statement about the given galvanic cell equation is -
Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
| 1. | The current will flow from silver to zinc in the external circuit. |
| 2. | The current will flow from zinc to silver in the external circuit. |
| 3. | The current will flow from silver to zinc in the internal circuit. |
| 4. | The current will flow from zinc to silver in the internal circuit. |
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Q. No.
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