1.

Given electrode potentials: Fe^(3+) + e^(-) rarr Fe^(2+) : E^(o) =0.771 V and I_(2) 2e^(-) rarr 2I^(-), E^(o) = 0.556 V then E^(o) for the cell reaction 2 Fe^(3+) 2I ^(-) rarr 2 Fe^(2+) + l_(2) will be:

Answer»

`(2XX 0.771 - 0. 536)= 1.006 ` VOLT
`(0.771 - 0.5 xx 0.536) = 0. 503` Volt
`0.771 – 0.536 = 0.235` Volt
`0.536 – 0.771 = – 0.235` Volt

Answer :C


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