Home
Class 12
CHEMISTRY
Given : Co^(3+)+e^(-)rarr Co^(2+), E^(...

Given :
`Co^(3+)+e^(-)rarr Co^(2+), E^(@)=+1.181V`
`Pb^(4+)+2e^(-)rarrPb^(2+),E^(@)=+1.67V`
`Ce^(4+)+e^(-) rarr Ce^(3+), E^(@)=+1.61V`
`Bi^(3+)+3e^(-)rarr Bi, E^(@)=+0.20V`
Oxidizing power of the species will increase in the order:

A

`Ce^(4+)lt Pb^(4+) lt Bi^(3+) lt Co`

B

`Co^(3+) lt Ce^(4+) lt Bi^(3+) lt Pb^(4+)`

C

`Bi^(3+) ltCe^(4+) lt Pb^(4+) lt Co^(3+)`

D

`Co^(3+) lt Pb^(4+) lt Ce^(4+) lt Bi^(3+)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the increasing order of oxidizing power of the given species based on their standard reduction potentials, we can follow these steps: ### Step 1: List the Given Reactions and Their Standard Reduction Potentials We have the following reactions and their corresponding standard reduction potentials (E°): 1. \( \text{Co}^{3+} + e^- \rightarrow \text{Co}^{2+}, \quad E^\circ = +1.181 \, \text{V} \) 2. \( \text{Pb}^{4+} + 2e^- \rightarrow \text{Pb}^{2+}, \quad E^\circ = +1.67 \, \text{V} \) 3. \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+}, \quad E^\circ = +1.61 \, \text{V} \) 4. \( \text{Bi}^{3+} + 3e^- \rightarrow \text{Bi}, \quad E^\circ = +0.20 \, \text{V} \) ### Step 2: Identify the Oxidizing Power The oxidizing power of a species is directly related to its standard reduction potential. A higher reduction potential indicates a stronger oxidizing agent. ### Step 3: Compare the Reduction Potentials Now, we will compare the given standard reduction potentials: - \( \text{Pb}^{4+} \) has the highest reduction potential of \( +1.67 \, \text{V} \) - \( \text{Ce}^{4+} \) follows with \( +1.61 \, \text{V} \) - \( \text{Co}^{3+} \) comes next with \( +1.181 \, \text{V} \) - \( \text{Bi}^{3+} \) has the lowest reduction potential of \( +0.20 \, \text{V} \) ### Step 4: Arrange in Increasing Order of Oxidizing Power Based on the reduction potentials, we can arrange the species in increasing order of their oxidizing power: 1. \( \text{Bi}^{3+} \) (lowest oxidizing power) 2. \( \text{Co}^{3+} \) 3. \( \text{Ce}^{4+} \) 4. \( \text{Pb}^{4+} \) (highest oxidizing power) ### Final Order Thus, the increasing order of oxidizing power of the species is: \[ \text{Bi}^{3+} < \text{Co}^{3+} < \text{Ce}^{4+} < \text{Pb}^{4+} \]

To determine the increasing order of oxidizing power of the given species based on their standard reduction potentials, we can follow these steps: ### Step 1: List the Given Reactions and Their Standard Reduction Potentials We have the following reactions and their corresponding standard reduction potentials (E°): 1. \( \text{Co}^{3+} + e^- \rightarrow \text{Co}^{2+}, \quad E^\circ = +1.181 \, \text{V} \) 2. \( \text{Pb}^{4+} + 2e^- \rightarrow \text{Pb}^{2+}, \quad E^\circ = +1.67 \, \text{V} \) 3. \( \text{Ce}^{4+} + e^- \rightarrow \text{Ce}^{3+}, \quad E^\circ = +1.61 \, \text{V} \) ...
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN REVISION TEST- 16

    VMC MODULES ENGLISH|Exercise CHEMISTRY (SECTION 2)|5 Videos
  • JEE MAIN REVISION TEST 8 (2020)

    VMC MODULES ENGLISH|Exercise CHEMISTRY (SECTION 2)|5 Videos
  • JEE MAIN REVISION TEST- 24

    VMC MODULES ENGLISH|Exercise CHEMISTRY (SECTION - 2)|5 Videos

Similar Questions

Explore conceptually related problems

Given : Co^(3+) + e^(-) to Co^(2+) , E^(@) = +1.81V pb^(4+) + 2e^(-) rightarrow Pb^(2+) , E^(@) = + 1.67V Ce^(4)+ e^(-)rightarrow Ce^(3+), E^(@) = +1.61V Ce^(4)+3e^(-)rightarrow Bi , E^(@) = + 0.20V Oxidation power of the species will increase in the order:

Half cell reactions for some electrodes are given below I. A + e^(-) rarr A^(-) , E^(@) = 0.96V II. B^(-) + e^(-) rarr B^(2-) , E^(@) = -0.12V III. C^(+) + e^(-) rarr C, E^(@) =+0.18V IV. D^(2+) + 2e^(-) rarr D, E^(@) = -1.12V Largest potential will be generated in which cell?

From the following E^(o) value of helf cells: (i) A+e^(-)rarr A ^(-) , E^(o) = -0.824 V (ii) B^(-)+e^(-)rarr B ^(2-) , E^(o) = +1.25V (iii) C^(-)+2e^(-)rarr C ^(3-) , E^(o) = -1.25V (iv) D^(-)+2e^(-)rarr D ^(2-) , E^(o) = +0.68V What combination of two half cells would result is a cell with the largest cell potential?

Find out E^(@) for F^(-) rarr (1)/(2)F_(2)+e^(-) , If F_(2)+2e^(-)rarr 2F^(-),E^(@)=+2.7V .

Standard electrode potential data is given below : Fe^(3+) (aq) + e ^(-) rarr Fe^(2+) (aq) , E^(o) = + 0.77 V Al^(3+) (aq) + 3e^(-) rarr Al (s) , E^(o) = - 1.66 V Br_(2) (aq) + 2e^(-) rarr 2Br^(-) (aq) , E^(o) = +1.08 V Based on the data given above, reducing power of Fe^(2+) Al and Br^(-) will increase in the order :

Cu^(+) + e rarr Cu, E^(@) = X_(1) volt, Cu^(2+) + 2e rarr Cu, E^(@) = X_(2) volt For Cu^(2+) + e rarr Cu^(+), E^(@) will be :

Cu^(+) + e rarr Cu, E^(@) = X_(1) volt, Cu^(2+) + 2e rarr Cu, E^(@) = X_(2) X_(2) volt For Cu^(2+) + e rarr Cu^(+), E^(@) will be :

Given that, Co^(3+) +e^(-)rarr Co^(2+) , E^(@) = +1.82 V 2H_(2)O rarr O_(2) +4H^(+) +4e^(-), E^(@) =- 1.23 V . Explain why Co^(3+) is not stable in aqueous solutions.

If {:(Sn^(2+) + 2e^(-) rarr Sn(s), E^(o) = - 0.14 V),(Sn^(4+)+2e^(-) rarr Sn^(2+),E^(o)= + 0.13 V):} then which of these is true?

Given: (i) Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V (ii) Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V Electrode potential, E^(@) for the reaction, Cu^(+)+e^(-) rarr Cu , will be