Question

Given electrode potentials asre
`Fe^(3+)+e^(-) rarr Fe^(2+)," "E^(c-)=0.771V`
I_(2)+2e^(-) rarr 2I^(c-)," "E^(c-)=0.536V`
`E^(c-)._(cell)` for the cell reaction,
`2Fe^(3+)+2I^(c-) rarr Fe^(2+)+I_(2)` is
A. `(2 xx 0.771 - 0.536 ) = 1.006` V
B. `(0.771 - 0.5 xx 0.536) = 0.503 V`
C. `0.771 - 0.536 = 0.235 V `
D. `0.536 - 0.771 = -0.236 V `

Answer 1

Correct Answer - C
`E_(cell)^(@) = E_(RHS)^(@) - E_(LHS)^(@) = E_("Higher" RP)^(@) - E_("Lower" RP)^(@)`
`=0.771 - 0.536 = +0.235` V