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E(cell)^(@) for the given redox reaction...

`E_(cell)^(@)` for the given redox reaction is 2.71 V
`Mg_((s))+Cu_((0.01 M))^(2+)toMg_((0.001 M))^(2+)+Cu_((s))`
Calculate `E_(cell)` for the reaction. Write the direction of flow of current when an external opposite potential applied is
(i) less than 2.71 V and (ii) greater than 2.71 V.

Text Solution

AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Understand the Reaction The given redox reaction is: \[ \text{Mg}_{(s)} + \text{Cu}^{2+}_{(0.01 M)} \rightarrow \text{Mg}^{2+}_{(0.001 M)} + \text{Cu}_{(s)} \] ### Step 2: Identify the Standard Cell Potential The standard cell potential \( E^\circ_{cell} \) is given as 2.71 V. ...
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Knowledge Check

  • For the cell reaction Zn_((s)) + Cu_("0.1 M")^(2+) to Zn_(0.1 M)^(2+) + Cu_((s)) if the standard EMF of the cell is E^(@) , then

    A
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    B
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    C
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    D
    `E le E^(@)`

  • For the redox reaction Zn(s) + Cu^(2+) (0.1M) rarr Zn^(2+) (1M) + Cu(s) that takes place in a cell, E^(o)""_(cell) is 1.10 volt. E_(cell) for the cell will be:

    A
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    B
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    C
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    D
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  • In a cell reaction, Cu_((s))+ 2Ag_((aq))^(+) to Cu_((aq))^(2+) + 2Ag_((s)) E_"cell"^@ =+0.46 V . If the concentration of Cu^(2+) ions is doubled then E_"cell"^@ will be

    A
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    B
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    C
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    D
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    Calculate E_(cell)^(@) for the following reaction at 298 K : 2 Cr (s ) + 3 Fe^(2+) (0.01 M) to 2 Cr^(3+) (0.01 M) + 3 Fe (s) Given : E_(cell) = 0.261 V

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