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The Edison storage cell is represented ...

The Edison storage cell is represented as `Fe_((s))// FeO_((s))// KOH_((aq))// Ni_2O_3// Ni_((s))`
The half - cell reaction are
`underset((s))(Ni_2O_3) + underset((l))(H_2O) + 2e^- to 2NiO + underset((aq))(2OH^-) , E^(@) = + 0.40V `
`underset((s)) + underset((l))(H_2O) + 2e^(-) to Fe(s) + underset((aq))(2OH^(-)), E^(@) = -0.87 V `
The cell potential is

A

`1.27 V `

B

`-0.47 V `

C

Cannot be determined, because it depends on `[OH^(-)]`

D

`E_("cell")^(@) + P^(OH)`

Text Solution

Verified by Experts

The correct Answer is:
A
(1) `DeltaG_1^0 = -2F (0.40)` (2) `DeltaG_2^0 = -2F (-0.87)`
`Fe_((s)) + Ni_2O_(3(s)) to FeO_((s)) + 2NiO_((s))`
`E_("cell")= E_(O.Pfe//FeO)^(0) - (0.059)/(2) log""([H_2O])/([OH]^2) + E_(R. PNi_2O_3 // NiO)^(0) + (0.059)/(2) log""([H_2O])/([OH]^(2))`
(1)` DeltaG_1^(0) = -2F (0.40) ` (2)` DeltaG_2^(0) = -2F(-0.87)`
(3) ` Fe + Ni_2O_3 to 2NiO + FeO , DeltaG_(3)^(0)= -2F(E_3^(0)),(3) = (1)-(2)`
` E_(3)^(0) = (E_1^(0)-E_2^(0)) = 0.40 - (-0.87) , E_3^(0) = + 1.27V`
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Knowledge Check

  • The Edison storage cell is represented as Fe_((s))// FeO_((s))// KOH_((aq))// Ni_2O_3// Ni_((s)) The half - cell reaction are underset((s))(Ni_2O_3) + underset((l))(H_2O) + 2e^- to 2NiO + underset((aq))(2OH^-) , E^(@) = + 0.40V underset((s)) + underset((l))(H_2O) + 2e^(-) to Fe(s) + underset((aq))(2OH^(-)), E^(@) = -0.87 V The maximum amount of electrical energy that can be obtained from one mole Ni_2O_3 .

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    `268.2 kJ`
    B
    `145.5 ` cal
    C
    `245.11 kJ`
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    `245. 11` cal

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