- 1(current)
Q. No.Pt (s) | \(H_2\) (g, 2 atm) | HCl (aq, 0.02 M)
\(E^0_{H_2 / H^+}\) = 0 V
(Given : \(\frac{2.303 RT}{F} = 0.059\),
log 2 = 0.3010)
1. 0.109 V
2. 0.035 V
3. − 0.035 V
4. − 0.109 V
Find the emf of the cell in which the following reaction takes place at 298 K:
\(\mathrm{Ni}(\mathrm{s})+2 \mathrm{Ag}^{+}(0.001 \mathrm{M}) \rightarrow \mathrm{Ni}^{2+}(0.001 \mathrm{M})+2 \mathrm{Ag}(\mathrm{s}) \)
\(\small{\text { (Given that } \mathrm{E}_{\text {cell }}^{\circ}=1.05 \mathrm{~V}; \dfrac{2.303 \mathrm{RT}}{\mathrm{F}}=0.059} )\)
| 1. | 1.05 V | 2. | 1.0385 V |
| 3. | 1.385 V | 4. | 0.9615 V |
Consider the given cell:
\(\mathrm{Z n \left|\right. Z n S O_{4} \left(\right. 0 . 01 M \left.\right) \left|\right. \left|\right. C u S O_{4} \left(\right. 1 . 0 M \left.\right) \left|\right. C u}\)
In the electrochemical cell, the emf of this Daniel cell is E1. When the concentration of ZnSO4 is changed to 1.0 M and that of CuSO4 is changed to 0.01 M, the emf changes to E2. From the following, which one is the relationship between E1 and E2 ?
(Given: \(\frac{RT}{F}\) = 0.059)
| 1. | \(\mathrm{E_{1} < E_{2}}\) | 2. | \(\mathrm{E_{1} > E_{2}}\) |
| 3. | \(\mathrm{E_{2} = 0 \neq E_{1}}\) | 4. | \(\mathrm{E_{1} = E_{2}}\) |
The pressure of H2 required to make the potential of H2 - electrode zero in pure water at 298 K is:
| 1. | 10–12 atm | 2. | 10–10 atm |
| 3. | 10–4 atm | 4. | 10–14 atm |
For a given reaction,
Cu(s) + 2Ag+ (aq) → Cu2+(aq)+2Ag(s);
E0=0.46 V at 298 K . The equilibrium constant will be :
| 1. | 2. | ||
| 3. | 4. |
A hypothetical electrochemical cell is shown below.
A|A+(x M) || B+(y M)|B
The Emf measured is +0.20 V. The cell reaction is:
1. A+ + B → A + B+
2. A+ + e- → A ; B+ + e- → B
3. The cell reaction cannot be predicted.
4. A + B+ → A+ + B
- 1(current)
Q. No.
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