Two half cell reactions are given below:
\(\begin{aligned} &\mathrm{{Co}^{3+}+e^{-} \rightarrow {Co}^{2+}, {E}_{{Co}^{2+} / {Co}^{3+}}^{\circ}=-1.81 {~V}} \\ &2 \mathrm{{Al}^{3+}+6 e^{-} \rightarrow 2 {Al}({s}), {E}_{{Al} / {Al}^{3+}}^{\circ}=+1.66 {~V}} \end{aligned} \)
The standard EMF of a cell with feasible redox reaction will be:
1. | +7.09 V | 2. | +0.15 V |
3. | +3.47 V | 4. | –3.47 V |
Consider the change in oxidation state of Bromine corresponding to different emf values as shown in the diagram below:
Then the species undergoing disproportionation is:-
1. | \(\text{BrO}^-_3\) | 2. | \(\text{BrO}^-_4\) |
3. | \(\text{Br}_2\) | 4. | \(\text{HBrO}\) |
In the electrochemical cell:
\(\mathrm{Z n \left|\right. Z n S O_{4} \left(\right. 0 . 01 M \left.\right) \left|\right. \left|\right. C u S O_{4} \left(\right. 1 . 0 M \left.\right) \left|\right. C u}, \)
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. \(\mathrm{E_{1} = E_{2}}\)
Zn2+(aq) + 2e–→ Zn(s) | Eo = – 0.76 V |
Ag2O(s) + H2O(l) + 2e– → 2Ag(s) + 2OH–(aq) | Eo = 0.34 V |
The cell potential will be:
1. | 0.42 V | 2. | 0.84 V |
3. | 1.34 V | 4. | 1.10 V |
1. | Y > X > Z | 2. | Z > X > Y |
3. | X > Y > Z | 4. | Y > Z > X |
Cu2+(aq) + e- → Cu+(aq) | 0.15 V |
Cu+(aq) + e- → Cu(s) | 0.50 V |
1. | 0.325 V | 2. | 0650 V |
3. | 0.150 V | 4. | 0.500 V |
Standard electrode potential for Sn4+/Sn2+ couple is +0.15 V and that for Cr3+/Cr couple is -0.74. These two couples in their standard state are connected to make a cell. The cell potential will be:
1. +0.89 V
2. +0.18 V
3. +1.83 V
4. +1.199 V
Consider the following relations for emf of an electrochemical cell:
(a) | emf of cell = (Oxidation potential of anode) – (Reduction potential of cathode) |
(b) | emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) |
(c) | emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) |
(d) | emf of cell = (Oxidation potential of anode) – (Oxidation potential of cathode) |
The correct relation among the given options is:
1. | (a) and (b) | 2. | (c) and (d) |
3. | (b) and (d) | 4. | (c) and (a) |