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Thus emf of the cell = 0.91 V. Q.3.6. Th...

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.

Text Solution

Verified by Experts

The reactions taking place in the cell may be written as under:
`2Fe^(3+) + 2e^(-) to 2Fe^(2+)` or `2I^(-) to I_(2) + 2e^(-)`
Thus, for the given cell reaction, n = 2. Using the following equation and substituting the values, we get
`Delta_(r)G^(@) =-nFE_("cell")^(@) =-2 xx 96500 xx 0.236 J = -45.55 kJ mol^(-1)`
Now, using the following relation to calculate cquilibrium constant
`Delta_(r)G^(@) = -2.303 RT log K_( C)`
or `K_( C) "antilog" (7.983) = 9.616 xx 10^(7)`
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