\(Fe^{2+}(aq) \ + \ Ag^{+}(aq) \ \rightarrow \ Fe^{3+}(aq) \ + \ Ag(s)\)
\(E_{Fe^{3+}/Fe^{2+}}^{o} \ = \ 0.77 \ V; \ E_{Ag^{+}/Ag}^{o} \ = \ 0.80 \ V\)
The value of \(\Delta G_{r}^{o} \) in the above reaction will be:
1. +
2.
3.
4. -186.83 J
Given:
(i) Eo = 0.337 V
(ii) Eo = 0.153 V
Electrode potential, Eo for the reaction,
, will be:
1. 0.52 V
2. 0.90 V
3. 0.30 V
4. 0.38 V
| 1. | K+> Na+> Rb+> Cs+ | 2. | Cs+> Rb+> K+> Na+ |
| 3. | Rb+> K+> Cs+> Na+ | 4. | Na+> K+> Rb+> Cs+ |
Given the following cell reaction:
\(\mathrm{2Fe^{3+}(aq) \ + \ 2I^{-}(aq)\rightarrow 2Fe^{2+}(aq) \ + \ I_{2}(aq)}\)
[Given: \(F = 96500\) \(C\) \(mol^{- 1}\)]
1. \(23 . 16\) \(kJ\) \(mol^{- 1}\)
2. \(- 46 . 32\) \(kJ\) \(mol^{- 1}\)
3. \(- 23 . 16\) \(kJ\) \(mol^{- 1}\)
4. \(46 . 32\) \(kJ\) \(mol^{- 1}\)
For a cell involving one electron \(E_{cell}^{\ominus} = 0 . 59 V\) at 298 K. The equilibrium constant for the cell reaction is :
\(\mathrm{[Given~ that~ \frac {2.303 ~RT}{F} = 0.059 ~V~ at~ T = 298 K]}\)
| 1. | \(1 . 0 \times \left(10\right)^{30}\) | 2. | \(1 . 0 \times \left(10\right)^{2}\) |
| 3. | \(1 . 0 \times \left(10\right)^{5}\) | 4. | \(1 . 0 \times \left(10\right)^{10}\)
|
| 1. | pH of the solution will rise. | 2. | pH of the solution will fall. |
| 3. | No change in the pH of the solution. | 4. | None of the above. |