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The equilibrium constant for the followi...

The equilibrium constant for the following reaction at 298K is expressed as `x xx10^(y)`
`2Fe^(3+)+2I^(-)to2Fe^(2+)+I_(2),E_(cell)^(@)=0.235V`
The value of y is.

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To solve the problem, we need to find the equilibrium constant \( K \) for the given reaction and express it in the form \( x \times 10^y \). The reaction is: \[ 2 \text{Fe}^{3+} + 2 \text{I}^- \rightarrow 2 \text{Fe}^{2+} + \text{I}_2 \] We are given the standard cell potential \( E^\circ_{\text{cell}} = 0.235 \, \text{V} \) at 298 K. ### Step 1: Determine the number of electrons transferred (n) ...
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The equilibrium constant of the following redox rection at 298 K is 1 xx 10^(8) 2Fe^(3+) (aq.) +2I^(-) (aq.) hArr 2Fe^(2+)(aq.) +I_(2) (s) If the standard reducing potential of iodine becoming iodide is +0.54 V. what is the standard reduction potential of Fe^(3+)//Fe^(2+) ?

Calculate the value of equilibrium constant for the reaction : 2Fe^(3+)+2I^(-) to 2Fe^(2+)+I_(2) Given that E_(cell)^(@)=0.235" V "

(a) The cell in which the following reactions occurs: 2Fe^(3+)(aq)=2I^(-)(aq)to2Fe^(2+)(aq)+I_(2)(s) has E_(cell)^(@)=0.236V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F=96,500" C "mol^(-1) ) (b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: 1F=96,500" C "mol^(-1) )