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Mg(s) + 2Ag^(+)(0.0001M) to 2Ag(s) + Mg^...

`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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Verified by Experts

`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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