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Using the data given below: E(Cr(2)O(7...

Using the data given below:
`E_(Cr_(2)O_(7)^(2-)|Cr^(3+))^(@)=1.33V E_(Cl_(2)|Cl^(-))^(@)=1.36V`
`E_(MnO_(4)^(-)|Mn^(2+))^(@)=1.51V E_(Cr^(3+)|Cr)=-0.74V`
Mark the strongest reducing agent.

A

`Cl^-`

B

`Cr`

C

`Cr^(3+)`

D

`Mn^(2+)`

Text Solution

AI Generated Solution

The correct Answer is:
To determine the strongest reducing agent from the given standard reduction potentials, we will follow these steps: ### Step 1: Understand the concept of reducing agents A reducing agent is a substance that donates electrons in a chemical reaction and gets oxidized in the process. The stronger the reducing agent, the higher its tendency to lose electrons. ### Step 2: Convert reduction potentials to oxidation potentials To find the strongest reducing agent, we need to convert the given standard reduction potentials (E°) into oxidation potentials. The oxidation potential is simply the negative of the reduction potential. ### Step 3: List the given reduction potentials We have the following reduction potentials: 1. \( E_{Cr_2O_7^{2-}/Cr^{3+}}^{\circ} = 1.33 \, V \) 2. \( E_{Cl_2/Cl^{-}}^{\circ} = 1.36 \, V \) 3. \( E_{MnO_4^{-}/Mn^{2+}}^{\circ} = 1.51 \, V \) 4. \( E_{Cr^{3+}/Cr}^{\circ} = -0.74 \, V \) ### Step 4: Calculate oxidation potentials Now, we will calculate the oxidation potentials for each half-reaction: 1. For \( Cr_2O_7^{2-}/Cr^{3+} \): \[ E_{ox} = -E_{red} = -1.33 \, V \] 2. For \( Cl_2/Cl^{-} \): \[ E_{ox} = -E_{red} = -1.36 \, V \] 3. For \( MnO_4^{-}/Mn^{2+} \): \[ E_{ox} = -E_{red} = -1.51 \, V \] 4. For \( Cr^{3+}/Cr \): \[ E_{ox} = -(-0.74 \, V) = 0.74 \, V \] ### Step 5: Compare oxidation potentials Now we have the following oxidation potentials: 1. \( Cr_2O_7^{2-}/Cr^{3+} \): -1.33 V 2. \( Cl_2/Cl^{-} \): -1.36 V 3. \( MnO_4^{-}/Mn^{2+} \): -1.51 V 4. \( Cr^{3+}/Cr \): 0.74 V ### Step 6: Identify the strongest reducing agent The strongest reducing agent will have the highest oxidation potential. From our calculations, the oxidation potential of \( Cr^{3+}/Cr \) is 0.74 V, which is the highest among all the calculated values. ### Conclusion Thus, the strongest reducing agent among the given species is **Chromium (Cr)**. ---

To determine the strongest reducing agent from the given standard reduction potentials, we will follow these steps: ### Step 1: Understand the concept of reducing agents A reducing agent is a substance that donates electrons in a chemical reaction and gets oxidized in the process. The stronger the reducing agent, the higher its tendency to lose electrons. ### Step 2: Convert reduction potentials to oxidation potentials To find the strongest reducing agent, we need to convert the given standard reduction potentials (E°) into oxidation potentials. The oxidation potential is simply the negative of the reduction potential. ...
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