1.

What is the potential for the cell `Cr|Cr^(3+)(0.1M)||Fe^(2+)(0.01M)|Fe` `E^(@)Cr^(3+)// Cr=-0.74V`, `E^(@)Fe^(2+)//Fe=-0.44V`

Answer» `2Cr(s) + 3Fe^(2+)(0.1M) rarr2Cr^(3+)(0.01M) + 3Fe(s)`
`E_("cell")=E_("cell")^(@)-(0.0591)/(n) log. ([Cr^(3+)]^(2))/([Fe^(2+)]^(3))`
Cell reaction :
`underset(("s"))(Cr) " | "underset(("0.01M")) (Cr^(3+))" || " underset(("0.01M"))(Fe^(2+))" | "Fe(s)`
`E_("cell")^(@)=E_("cathode")^(@)-E_("amode")^(@)`
`=E^(@)(Fe^(2+)|Fe) - E^(@) (Cr^(3+)|Cr)`
= `- 0.44 - (-0.74)`
`E_("cell")^(@) = - 0.44 + 0.74 = 0.30V`
`n = 6e^(-)`
`E_("cell")^(@) = 0.30 - (0.0591)/(6)log.((0.01)^(2))/((0.1)^(3)) = 0.30 - (0.0591)/(6) log. 10^(-1)`
`0.30 - (0.0591)/(6) xx (-1) = 0.30 + (0.0591)/(6)`
=0.30 + 0.00985 = 0.309V


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