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Mg(s) + 2Ag^(+)(0.0001M) to 2Ag(s) + Mg^(2+) (0.13 M). Calculate the E_("cell") is given as 3.17 V. |
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Answer» Solution :`E_("cell")` is written DIRECTLY from `E_("cell")^(@)` as, `E_("cell") = E_("cell")^(@) - (RT)/(2F) "LN" ([Mg^(2+)])/([Ag^+]^(2)) "(or)" E_("cell") = 3.17 - (0.059)/2 "log" (0.13)/((0.0001)^(2))` `E_("cell")` of the GIVEN reaction = `3.17 - 0.21 = 2.96 V`. |
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