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The cell in which the following reaction...

The cell in which the following reaction occurs
`2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)++I_(2)(s)` has `E_(cell)^(0)=0.236V` at 298 K.
Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Text Solution

AI Generated Solution

To solve the given problem, we need to calculate the standard Gibbs free energy change (\(\Delta G^\circ\)) and the equilibrium constant (\(K_c\)) for the cell reaction: \[ 2Fe^{3+}(aq) + 2I^-(aq) \rightarrow 2Fe^{2+}(aq) + I_2(s) \] Given data: - Standard cell potential, \(E^\circ_{cell} = 0.236 \, V\) - Temperature, \(T = 298 \, K\) ...
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The cell in which the following reaction occurs: 2Fe^(3+)(aq)+2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Thus emf of the cell = 0.91 V. Q.3.6. The cell in which the following reaction occurs : 2Fe^(3+) (aq) + 2I^(-)(aq) to 2Fe^(2+0 (aq) + I_(2)(s) has E_("cell")^(@) = 0.236 V at 298 K Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.

Knowledge Check

  • The cell in which the following reaction occurs : 2Fe_(aq)^(3+) + 2I_(aq)^(-) to 2Fe_(aq)^(2+) + I_(2(s)) "has" E_("cell")^(o) = 0.236 V "at" 298 K The equilibrium constnat of the cell reaction is

    A
    `6.69xx 10^(-7)`
    B
    `9.69 xx 10^(-7)`
    C
    `9.69 xx 10^(7)`
    D
    `6.69 xx 10^(7)`

  • For the cell reaction: 2Fe^(3+)(aq)+2l^(-)(aq)to2Fe^(2+)(aq)+l_(2)(aq) E_(cell)^(ɵ)=0.24V at 298K . The rstandard gibbs energy (triangle,G^(ɵ)) of the cell reaction is [Given that faraday constnat F=96400Cmol^(-1)]

    A
    `23.16kJmol^(-1)`
    B
    `-46.32kJmol^(-1)`
    C
    `-23.16kJmol^(-1)`
    D
    `46.32kJmol^(-1)`

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    The cell in which the following reaction occurs 2Fe^(3+)(aq)+2I^(-)(aq) to 2Fe^(2+)(aq)+2I_(2) " has "E_(cell)^(@)=0.236 V " at "298 K . Calculate standard Gibbs energy and equilibrium constant for the reaction.

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