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Zn|Zn^(2+)(a= 0.1M) || Fe^(2+) (a=0.01M)...

`Zn|Zn^(2+)(a= 0.1M) || Fe^(2+) (a=0.01M)|Fe`
The emf of the above cell is 0.2905 V. Equilibrium constant for the cell reaction is:

A

`10^(0.32//0.059)`

B

`10^(0.32//0.0295)`

C

`10^(0.26//0.0295)`

D

`10^(0.32//0.295)`

Text Solution

AI Generated Solution

To find the equilibrium constant for the given electrochemical cell reaction, we will follow these steps: ### Step 1: Identify the half-reactions and the overall cell reaction The cell notation provided is: \[ \text{Zn} | \text{Zn}^{2+} (a = 0.1M) || \text{Fe}^{2+} (a = 0.01M) | \text{Fe} \] From this notation, we can identify the half-reactions: - Oxidation at the anode: ...
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The standard e.m.f. of the cell Zn| Zn^(2+) (0.01 M) || Fe^(2+) (0.001 M) | Fe at 298 K is 0.02905 then the value of equilibrium constant for the cell reaction 10^(x) . Find x.

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Knowledge Check

  • Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)|Fe . The emf of the above cell is 0.2905V. Equilibrium constant for the cell reaction is

    A
    `10^(0.32//0.059)`
    B
    `10^(0.32//0.0295)`
    C
    `10^(0.26//0.0295)`
    D
    `10^(0.32//0.295)`

  • Zn|Zn^(2+)(a=0.1 M)||Fe^(2+)(a=0.01M)|Fe the emf of the above cell is 0.2905 V Equilibrium constant for the cell reaction is

    A
    `10^(0.32//0.591)`
    B
    `10^(0.32//0.0295)`
    C
    `10^(0.26//0.295)`
    D
    `10^(0.32//0.295)`

  • Zn|Zn^(2+)(a=0.1M)||Fe^(2+)(a=0.01M)Fe. The EMF of the above cell is 0.2905 . The equilibrium constant for the cell reaction is

    A
    `10^(0.32//0.0591)`
    B
    `10^(0.32//0.0295)`
    C
    `10^(0.26//0.0295)`
    D
    `e^(0.32//0.2995)`

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